Tutors Answer Your Questions about Finance (FREE)
Question 1167294: The supply and demand for a product are given by
2p − q = 60 and pq = 100 + 25q,
respectively. Find the market equilibrium point.
p,q=
Answer by Resolver123(6)  (Show Source):
You can put this solution on YOUR website! We are given the supply and demand equations:
Supply equation: 2p - q = 60 ...................(1)
Demand equation: pq = 100 + 25q ................(2)
The market equilibrium point occurs where supply = demand, which means we need to find the values of p and q that satisfy both equations simultaneously.
Solve equation (1) for p:
From the supply equation:
2p - q = 60 => 2p = 60 + q =>  ........(3)
Substitute (3) into (2):
 , or 
Rearrange the terms into a quadratic equation
 , or  .
The left side of the quadratic equation can be factored as  , giving q = -20 or q = 10.
Eliminate q = -20 since it is negative. Therefore, q = 10. So the equilibrium quantity is q = 10.
Substitute q = 10 into equation (3) to find the corresponding price:
 . Hence, the equilibrium price is p = 35.
Thus, the market equilibrium point is (q, p) = (10, 35), which means that the market is in equilibrium when 10 units are sold at a price of $35.
Question 1167311: tim did 10 lunges on day 1 and continued this doing 4 more lunges on each day.on day 7th he took a breakand did nt do any day day.what is the equation for this? express it in the form y=ax(x+b)+c, where a,b and c are constants.x= no of days and y=no of lunges timid
Answer by ikleyn(52781)  (Show Source):
Question 1167337: tim did 10 lunges on day 1 and continued this doing 4 more lunges on each day.on day 7th he took a breakand did nt do any day day.what is the equation for this? express it in the form y=ax(x+b)+c, where a,b and c are constants.x= no of days and y=no of lunges tim Did
Answer by ikleyn(52781) (Show Source):
Question 1206514: If you make quarterly deposits of $407
into an ordinary annuity earning an annual interest rate of 6.52%
, how much will be in the account after 8
years? How much interest did you earn in those 8
years?
How much is in the account after 8
years?
How much interest was earned after
8 years?
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
If you make quarterly deposits of $407 into an ordinary annuity earning an annual interest rate of 6.52%,
how much will be in the account after 8 years? How much interest did you earn in 8 years?
~~~~~~~~~~~~~~~~~~~~~~~~~
Use a standard formula for the future value of an ordinary annuity
FV =  .
Here 'D' is the quarterly deposit of $407; 'r' is the effective interest rate
per quarter r = 0.062/4; 'n' is the number of deposits n = 4*8 = 32 (same as the number of quarters.
Thus you have
FV =  = 16,920.49.
This is the value at the account in 8 years.
You deposited 8*4*407 = 13024 dollars.
The interest is the difference 16,920.49 - 13024 = 3896.49 dollars.
Solved.
It is all what you need to know and all what you need to do to solve this problem.
Question 1165913: . James has $3,000 in credit card debt, which charges 14% interest. How long will it take to
pay off the card if he makes the minimum payment of $60 a month?
Answer by ikleyn(52781) (Show Source):
Question 1166516: Suppose you invest $120 a month for 3 years into an account earning 8% compounded monthly. After 3 years, you leave the money, without making additional deposits, in the account for another 28 years. How much will you have in the end?
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
a TWIN problem was solved at this forum under this link
https://www.algebra.com/algebra/homework/Finance/Finance.faq.question.1166134.html
Enjoy !
Question 1165476: The volume of a truncated prism with an equilateral triangle as its horizontal base is equal to 3600 cm3. The vertical edges at
each corner are 4, 6, and 8 cm. respectively. Find one side of the base.
a. 22.37 b. 25.43 c. 37.22 d. 17.89
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
The volume of a truncated prism with an equilateral triangle as its horizontal base is equal to 3600 cm3.
The vertical edges at each corner are 4, 6, and 8 cm. respectively. Find one side of the base.
a. 22.37 b. 25.43 c. 37.22 d. 17.89
~~~~~~~~~~~~~~~~~~~~~~~~
First, let's look at the position of the point at the sloped section,
which is exactly over the center of the regular triangle at the base.
Vertical coordinate of this point is the arithmetic mean of the numbers 4, 6 and 8
z =  = 6 cm over the base plane.
Now, let's accept without a proof, that the volume of the truncated prism is the same
as the volume of the regular prism with the same base and the height of 6 sm.
If the side of the base is 'a', then the area of the base is  cm^2,
and the volume of the regular prism is  cm^3 =  cm^3.
It gives us an equation for 'a'
= 3600.
From this equation
 =  =  =  =  ,
a =  = 37.22 cm (rounded).
ANSWER. The size of the base is about 37.22 cm , which is option (c) in the answer list.
Solved.
Question 1166134: Suppose you invest $120 a month for 6 years into an account earning 8% compounded monthly. After 6 years, you leave the money, without making additional deposits, in the account for another 20 years. How much will you have in the end?
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
Suppose you invest $120 a month for 6 years into an account earning 8% compounded monthly.
After 6 years, you leave the money, without making additional deposits, in the account for another 20 years.
How much will you have in the end?
~~~~~~~~~~~~~~~~~~~~~~~~~
For the first 6 years, it works as an ordinary annuity, and creates Future value of
FV =  = 11043.04 (rounded to the closest cent).
In the next 20 years, it works as an one-time deposit of 11043.04 compounded monthly at 8%.
Working this way, it produces the Future value / (final) amount of
 = 54406.88.
ANSWER. The final value is $54406.88.
Solved.
Question 1165581: How long must a temporary warehouse last to be desirable investment if it costs $ 16,000 to build, has annual maintenance and
operating cost of $ 360, provides storage space-value at $ 3,600 per year and if the company MARR is 10%? Hint: Treat MARR
as interest rate, i.
a. 8 years b. 9 years c.7 years d. 6 years
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
How long must a temporary warehouse last to be desirable investment if it costs $ 16,000 to build,
has annual maintenance and operating cost of $ 360, provides storage space-value at $ 3,600 per year
and if the company MARR is 10%?
Hint: Treat MARR as interest rate, i.
a. 8 years b. 9 years c.7 years d. 6 years
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
First, the definition of the MARR conception from Google AI:
In finance, MARR stands for Minimum Acceptable Rate of Return, also known as the hurdle rate
or minimum attractive rate of return. It is the minimal profit that an investor or company
is willing to accept from an investment or project, considering the project's risk and
the opportunity cost of other available investments.
A project must exceed its MARR to be considered financially viable and worthwhile for investment.
Now we can re-phrase the problem in more accessible Math terms, that are understandable
not only to specialists in Finance, but also to a wider audience of regular students.
A company "Warehouse, Inc" is going to loan $16,000 to construct a warehouse.
The warehouse will provide the storage space-value at $ 3,600 per year. It will have the annual
maintenance and operating cost of $360 per year. The loan interest is 10% compounded yearly.
How long the warehouse must function to recoup the cost of the loan and other relevant costs?
Solution
Let PMT be the annual payment for the loan (which is some constant value over the years).
Then the inequality for the project to be profitable is
PMT + 360 <= 3600, (1)
which implies
PMT <= 3600 - 360 = 3240. (2)
Now, let's calculate PMT for 6, 7, 8 and 9 years with the interest rate r = 10% = 0.1.
Use the standard formula
for n = 6 years PMT =  =  = 3673.72;
for n = 7 years PMT = =  = 3286.49;
for n = 8 years PMT = =  = 2999.10;
for n = 9 years PMT = =  = 2778.25.
Comparing these numbers with 3240 in the right side of (2), you see that to be profitable,
the warehouse should function at least 8 years. <<<---=== ANSWER
Solved.
Question 1209539: Gabriella and Mario plan to send their son to university. To pay for this they will contribute 9 equal yearly payments to an account bearing interest at the APR of 3.4%, compounded annually. Five years after their last contribution, they will begin the first of five, yearly, withdrawals of $35,600 to pay the university's bills. How large must their yearly contributions be?
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
In the post by @ElectricPavlov, the logic of the solution is correct.
The formulas that he uses for calculations are correct, too.
But all his output numbers, that are the results of his calculations, are WRONG.
His final answer is wrong, too.
It is because they systematically use inadequate calculator or inadequate calculation procedure
for their calculations, which does not provide the necessary precision.
So, if you are looking to get a precise answer for your problems in Finance,
which would be correct at the reference level of accuracy for your answer book,
especially for annuity problems, then @ElectricPavlov is a bad source for such a goal.
As I look at their performance, I clearly see that they do not understand, at all,
what approximate calculations are and which requirements such calculations must satisfy.
So, their level of understanding approximate calculations corresponds to a 5th grade school student,
who makes such calculations for the first time in his life and who never took/got lessons
from an experienced mentor on the subject.
Question 1194896: Please help with the homework:
Calculate the sum accrued on a fixed deposit of R10000 is invested on 15 march 2021 until 1 July 2023, if interest is credited annually on the 1 July at 15,5%.
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
Calculations in the post by @ElectricPavlov are incorrect.
For example, he calculates 1.155^2.2986 and gets approximately 1.404,
which leads him to the answer 14,040.
Correct calculation is 1.155^2.2986 = 1.39267875, giving the final answer 13,926.79.
So, the answer R14,040 in the post by @ElectricPavlov is incorrect.
Also, from the problem, it is unclear, if the interest is simple or compounded,
so, the problem itself is dark and its meaning is unclear.
As the problem is worded and presented in the post, it is as far from to be
a true Math problem as the ground is far from the heaven.
Question 1193499: Julita owes Jose the following amounts:
a. 55,000 at the end of 4 years,
b. 48,000 at the end of 5 years, and
c. 75,000 due in 3 years from today at 9% converted quarterly.
What single payment at the end of 2 years will settle Julita’s
obligations if money is worth 15% converted monthly?
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
Julita owes Jose the following amounts:
a. 55,000 at the end of 4 years,
b. 48,000 at the end of 5 years, and
c. 75,000 due in 3 years from today at 9% converted quarterly.
What single payment at the end of 2 years will settle Julita’s
obligations if money is worth 15% converted monthly?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I checked calculations in the post by @ElectricPavlov.
They ALL are INCORRECT.
The general formulas are correct, but each and every calculation in his post produces WRONG NUMBER as an output.
In addition, the problem's formulation in the post is MONSTROUS: its different parts
are not consistent each to other, so at these circumstances it is hopeless even to try to solve
the problem: firstly it should be formulated/worded/presented in a right way.
As it is presented in the post, it is not a Math problem.
It is soup of words, which is as far from to be a Math problem, as the ground is far from the heaven.
Question 1167763: Assume the car can be purchased for 0% down for 60 months (in lieu of rebate).
A car with a sticker price of $42,150 with factory and dealer rebates of $5,100
(a) Find the monthly payment if financed for 60 months at 0% APR. (Round your answer to the nearest cent.)
$
(b) Find the monthly payment if financed at 2.5% add-on interest for 60 months. (Round your answer to the nearest cent.)
$
(c) Use the APR approximation formula to find the APR for part (b). (Round your answer to one decimal place.)
%
(d) State whether the 0% APR or the 2.5% add-on rate should be preferred.
0% APR
2.5% add-on rate
Answer by CPhill(1959)  (Show Source):
You can put this solution on YOUR website! ```python?code_reference&code_event_index=2
# Part (a): Monthly payment with 0% APR
sticker_price = 42150
rebates = 5100
loan_amount = sticker_price - rebates
loan_term_months = 60
monthly_payment_0_apr = loan_amount / loan_term_months
print(f"Monthly payment (0% APR): {monthly_payment_0_apr:.2f}")
# Part (b): Monthly payment with 2.5% add-on interest
add_on_rate_annual = 0.025
loan_term_years = loan_term_months / 12
total_add_on_interest = loan_amount * add_on_rate_annual * loan_term_years
total_amount_to_repay = loan_amount + total_add_on_interest
monthly_payment_add_on = total_amount_to_repay / loan_term_months
print(f"Total add-on interest: {total_add_on_interest:.2f}")
print(f"Total amount to repay: {total_amount_to_repay:.2f}")
print(f"Monthly payment (2.5% add-on interest): {monthly_payment_add_on:.2f}")
# Part (c): APR for part (b) using the APR approximation formula
n = 12 # number of payment periods in one year
I = total_add_on_interest
P = loan_amount
N = loan_term_months
apr_approximation = (2 * n * I) / (P * (N + 1))
apr_percentage = apr_approximation * 100
print(f"APR approximation: {apr_percentage:.1f}%")
```
```text?code_stdout&code_event_index=2
Monthly payment (0% APR): 617.50
Total add-on interest: 4631.25
Total amount to repay: 41681.25
Monthly payment (2.5% add-on interest): 694.69
APR approximation: 4.9%
```
Here are the calculations for the car purchase:
**Loan Amount:**
Sticker price: $$42,150$
Factory and dealer rebates: $$5,100$
Loan Amount = $$42,150 - $5,100 = $37,050$
**(a) Find the monthly payment if financed for 60 months at 0% APR.**
Since the APR is 0%, there is no interest charged.
Monthly Payment = Loan Amount / Number of Months
Monthly Payment = $$37,050 / 60 = $617.50$
Rounding to the nearest cent, the monthly payment is $\\boxed{$617.50}$.
**(b) Find the monthly payment if financed at 2.5% add-on interest for 60 months.**
Loan Amount = $$37,050$
Add-on Interest Rate = 2.5% per year
Loan Term = 60 months = 5 years
Total Add-on Interest = Loan Amount $\\times$ Add-on Interest Rate $\\times$ Number of Years
Total Add-on Interest = $$37,050 \\times 0.025 \\times 5 = $4,631.25$
Total Amount to Repay = Loan Amount + Total Add-on Interest
Total Amount to Repay = $$37,050 + $4,631.25 = $41,681.25$
Monthly Payment = Total Amount to Repay / Number of Months
Monthly Payment = $$41,681.25 / 60 = $694.6875$
Rounding to the nearest cent, the monthly payment is $\\boxed{$694.69}$.
**(c) Use the APR approximation formula to find the APR for part (b).**
APR Approximation Formula: $APR \\approx \\frac{2 \\times n \\times I}{P \\times (N+1)}$
Where:
$n$ = number of payment periods in one year = 12 (for monthly payments)
$I$ = total interest paid = $$4,631.25$ (from part b)
$P$ = principal (loan amount) = $$37,050$
$N$ = total number of payments = 60
$APR \\approx \\frac{2 \\times 12 \\times $4,631.25}{$37,050 \\times (60+1)}$
$APR \\approx \\frac{24 \\times $4,631.25}{$37,050 \\times 61}$
$APR \\approx \\frac{$111,150}{$2,260,050}$
$APR \\approx 0.04918$
Convert to percentage and round to one decimal place:
$APR \\approx 0.04918 \\times 100% \\approx 4.918%$
Rounding to one decimal place, the APR is $\\boxed{4.9%}$.
**(d) State whether the 0% APR or the 2.5% add-on rate should be preferred.**
Comparing the two options:
* 0% APR: Monthly payment of $$617.50$ with no interest.
* 2.5% add-on rate: Monthly payment of $$694.69$ with an effective APR of approximately 4.9%.
The 0% APR option results in a lower monthly payment and incurs no interest, making it clearly the better financial choice.
The 0% APR should be preferred.
Question 1167857: A company buys a policy to insure its revenue in the event of major snowstorms that shut down business. The policy pays nothing for the first such snowstorm of the year and $10600 for each one thereafter, until the end of the year. The number of major snowstorms per year that shut down business has a Poisson distribution with mean 1.8.
Find the expected amount paid to the company under this policy during a one-year period.
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let $N$ be the number of major snowstorms per year that shut down business.
$N$ follows a Poisson distribution with mean $\lambda = 1.8$.
The probability mass function (PMF) is $P(N=k) = \frac{e^{-\lambda} \lambda^k}{k!}$ for $k = 0, 1, 2, \dots$.
Let $X$ be the amount paid to the company under the policy.
The policy pays nothing for the first snowstorm.
The policy pays $\$10600$ for each one thereafter.
We can define $X$ as follows:
* If $N=0$ (no snowstorms), $X=0$.
* If $N=1$ (one snowstorm), $X=0$ (the first one is free).
* If $N=k$ for $k \ge 2$, the number of payable snowstorms is $k-1$. So, $X = 10600(k-1)$.
We want to find the expected amount paid, $E[X]$.
$E[X] = \sum_{k=0}^{\infty} X(k) P(N=k)$
$E[X] = 0 \cdot P(N=0) + 0 \cdot P(N=1) + \sum_{k=2}^{\infty} 10600(k-1) P(N=k)$
$E[X] = 10600 \sum_{k=2}^{\infty} (k-1) \frac{e^{-\lambda} \lambda^k}{k!}$
We can factor out $e^{-\lambda}$:
$E[X] = 10600 e^{-\lambda} \sum_{k=2}^{\infty} (k-1) \frac{\lambda^k}{k!}$
Let's evaluate the sum $S = \sum_{k=2}^{\infty} (k-1) \frac{\lambda^k}{k!}$:
$S = (2-1)\frac{\lambda^2}{2!} + (3-1)\frac{\lambda^3}{3!} + (4-1)\frac{\lambda^4}{4!} + \dots$
$S = \frac{\lambda^2}{2!} + \frac{2\lambda^3}{3!} + \frac{3\lambda^4}{4!} + \dots$
We can use the property that $\frac{k-1}{k!} = \frac{k}{k!} - \frac{1}{k!} = \frac{1}{(k-1)!} - \frac{1}{k!}$.
So, $S = \sum_{k=2}^{\infty} \left( \frac{1}{(k-1)!} - \frac{1}{k!} \right) \lambda^k$
$S = \sum_{k=2}^{\infty} \frac{\lambda^k}{(k-1)!} - \sum_{k=2}^{\infty} \frac{\lambda^k}{k!}$
For the first part of the sum: $\sum_{k=2}^{\infty} \frac{\lambda^k}{(k-1)!} = \lambda \sum_{k=2}^{\infty} \frac{\lambda^{k-1}}{(k-1)!}$
Let $j=k-1$. When $k=2$, $j=1$.
$= \lambda \sum_{j=1}^{\infty} \frac{\lambda^j}{j!} = \lambda \left( \frac{\lambda^1}{1!} + \frac{\lambda^2}{2!} + \dots \right)$
We know the Taylor series for $e^\lambda$ is $\sum_{j=0}^{\infty} \frac{\lambda^j}{j!} = 1 + \frac{\lambda^1}{1!} + \frac{\lambda^2}{2!} + \dots = e^\lambda$.
So, $\sum_{j=1}^{\infty} \frac{\lambda^j}{j!} = e^\lambda - 1$.
Thus, the first part is $\lambda(e^\lambda - 1)$.
For the second part of the sum: $\sum_{k=2}^{\infty} \frac{\lambda^k}{k!}$
This is the Taylor series for $e^\lambda$ minus the first two terms ($k=0$ and $k=1$):
$\sum_{k=2}^{\infty} \frac{\lambda^k}{k!} = \left( \sum_{k=0}^{\infty} \frac{\lambda^k}{k!} \right) - \frac{\lambda^0}{0!} - \frac{\lambda^1}{1!} = e^\lambda - 1 - \lambda$.
Now substitute these back into $S$:
$S = \lambda(e^\lambda - 1) - (e^\lambda - 1 - \lambda)$
$S = \lambda e^\lambda - \lambda - e^\lambda + 1 + \lambda$
$S = \lambda e^\lambda - e^\lambda + 1$
Finally, multiply by $e^{-\lambda}$ to get $E[X]$:
$E[X] = 10600 e^{-\lambda} (\lambda e^\lambda - e^\lambda + 1)$
$E[X] = 10600 (\lambda - 1 + e^{-\lambda})$
Now, substitute the given value $\lambda = 1.8$:
$E[X] = 10600 (1.8 - 1 + e^{-1.8})$
$E[X] = 10600 (0.8 + e^{-1.8})$
Using a calculator, $e^{-1.8} \approx 0.1652988$
$E[X] = 10600 (0.8 + 0.1652988)$
$E[X] = 10600 (0.9652988)$
$E[X] \approx 10232.167288$
Rounding to two decimal places for currency:
$E[X] \approx \$10232.17$
The final answer is $\boxed{10232.17}$.
Question 1167948: Assume the car can be purchased for 0% down for 60 months (in lieu of rebate).
A BMW that has a sticker price of $62,590 with factory and dealer rebates of $6,000
(a) Find the monthly payment if financed for 60 months at 0% APR. (Round your answer to the nearest cent.)
$
(b) Find the monthly payment if financed at 2.5% add-on interest for 60 months. (Round your answer to the nearest cent.)
$
(c) Use the APR approximation formula to find the APR for part (b). (Round your answer to one decimal place.)
%
(d) State whether the 0% APR or the 2.5% add-on rate should be preferred.
0% APR
2.5% add-on rate
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! ```python?code_reference&code_event_index=2
# Part (a): Monthly payment with 0% APR
sticker_price = 62590
rebates = 6000
loan_amount = sticker_price - rebates
loan_term_months = 60
monthly_payment_0_apr = loan_amount / loan_term_months
print(f"Monthly payment (0% APR): {monthly_payment_0_apr:.2f}")
# Part (b): Monthly payment with 2.5% add-on interest
add_on_rate_annual = 0.025
loan_term_years = loan_term_months / 12
total_add_on_interest = loan_amount * add_on_rate_annual * loan_term_years
total_amount_to_repay = loan_amount + total_add_on_interest
monthly_payment_add_on = total_amount_to_repay / loan_term_months
print(f"Total add-on interest: {total_add_on_interest:.2f}")
print(f"Total amount to repay: {total_amount_to_repay:.2f}")
print(f"Monthly payment (2.5% add-on interest): {monthly_payment_add_on:.2f}")
# Part (c): APR for part (b) using the APR approximation formula
n = 12 # number of payment periods in one year
I = total_add_on_interest
P = loan_amount
N = loan_term_months
apr_approximation = (2 * n * I) / (P * (N + 1))
apr_percentage = apr_approximation * 100
print(f"APR approximation: {apr_percentage:.1f}%")
```
```text?code_stdout&code_event_index=2
Monthly payment (0% APR): 943.17
Total add-on interest: 7073.75
Total amount to repay: 63663.75
Monthly payment (2.5% add-on interest): 1061.06
APR approximation: 4.9%
```
Here are the calculations for the car purchase:
**Loan Amount:**
Sticker price: $$62,590$
Factory and dealer rebates: $$6,000$
Loan Amount = $$62,590 - $6,000 = $56,590$
**(a) Find the monthly payment if financed for 60 months at 0% APR.**
Since the APR is 0%, there is no interest charged.
Monthly Payment = Loan Amount / Number of Months
Monthly Payment = $$56,590 / 60 = $943.1666...$
Rounding to the nearest cent, the monthly payment is $\\boxed{$943.17}$.
**(b) Find the monthly payment if financed at 2.5% add-on interest for 60 months.**
Loan Amount = $$56,590$
Add-on Interest Rate = 2.5% per year
Loan Term = 60 months = 5 years
Total Add-on Interest = Loan Amount $\\times$ Add-on Interest Rate $\\times$ Number of Years
Total Add-on Interest = $$56,590 \\times 0.025 \\times 5 = $7,073.75$
Total Amount to Repay = Loan Amount + Total Add-on Interest
Total Amount to Repay = $$56,590 + $7,073.75 = $63,663.75$
Monthly Payment = Total Amount to Repay / Number of Months
Monthly Payment = $$63,663.75 / 60 = $1,061.0625$
Rounding to the nearest cent, the monthly payment is $\\boxed{$1,061.06}$.
**(c) Use the APR approximation formula to find the APR for part (b).**
APR Approximation Formula: $APR \\approx \\frac{2 \\times n \\times I}{P \\times (N+1)}$
Where:
$n$ = number of payment periods in one year = 12 (for monthly payments)
$I$ = total interest paid = $$7,073.75$ (from part b)
$P$ = principal (loan amount) = $$56,590$
$N$ = total number of payments = 60
$APR \\approx \\frac{2 \\times 12 \\times $7,073.75}{$56,590 \\times (60+1)}$
$APR \\approx \\frac{24 \\times $7,073.75}{$56,590 \\times 61}$
$APR \\approx \\frac{$169,770}{$3,452,990}$
$APR \\approx 0.04916$
Convert to percentage and round to one decimal place:
$APR \\approx 0.04916 \\times 100% \\approx 4.916%$
Rounding to one decimal place, the APR is $\\boxed{4.9%}$.
**(d) State whether the 0% APR or the 2.5% add-on rate should be preferred.**
Comparing the two options:
* 0% APR: Monthly payment of $$943.17$ with no interest.
* 2.5% add-on rate: Monthly payment of $$1,061.06$ with an effective APR of approximately 4.9%.
The 0% APR option results in a lower monthly payment and incurs no interest, making it clearly the better financial choice.
The 0% APR should be preferred.
Question 1164621: Suppose that A is a square matrix and A4 = 0 (the zero matrix). Show that (10 points)
(I − A)
−1 = I + A + A^2 + A^3
.
Answer by ikleyn(52781) (Show Source):
Question 1166885: Without expanding find the constant term and term involving x^6 in the expansion of (4x-(1/3)x²)^18
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
(a) Without expanding find the constant term in the expansion of (4x-(1/3)x²)^18.
(b) Without expanding find the term involving x^6 in the expansion of (4x-(1/3)x²)^18.
~~~~~~~~~~~~~~~~~~~~~~~~~~~
(a) The general formula for the k-th term of this expansion is
 =  , k = 0, 1, 2, 3, . . . , 18.
The constant term is the term with (18-k) + 2k = 0, or 18 + k = 0, k = -18.
But this value, k= -18 is not in the range [0,18] of possible values of 'k'.
So, we conclude that the constant term absents in this expansion.
It is the same as to say that the constant term has the coefficient '0' in the expansion.
It becomes obvious as you look at the basic binomial: you will see that x does present
in both terms of the binomial; so, the constant term, independent of x, absents in the expansion.
(b) Again, the general formula for the k-th term of this expansion is
= , k = 0, 1, 2, 3, . . . , 18.
The term involving x^6 is the term with (18-k) + 2k = 6, or 18 + k = 6, k = 6 - 18 = -12.
But this value, k= -12 is not in the range [0,18] of possible values of 'k'.
So, we conclude that the constant term absents in this expansion.
It is the same as to say that the term involving x^6 has the coefficient '0' in the expansion.
It becomes obvious as you look at the basic binomial: you will see that x does present
in both terms of the binomial; so, the degrees of 'x' in the expansion all are at least 18.
Solved.
Question 1168530: A regular octagon is emclosed with a circle.it has an extension piece CDFE obtained by extendind F and C to E with BCE is a straight line.AF was the diameter of the circle. FE is a tangent to the circle at F.calclate the size of z(angle FEC)
Found 2 solutions by ikleyn, CPhill:
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
A regular octagon is emclosed with a circle.it has an extension piece CDFE obtained by extendind
F and C to E with BCE is a straight line.AF was the diameter of the circle. FE is a tangent to the circle at F.
calclate the size of z(angle FEC)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The way how the problem is presented in the post, is  .
It is not a way presenting Math problem.
Throw this composition to a garbage bin, where its right place is.
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's break down this geometry problem step by step.
**1. Diagram and Definitions**
* We have a regular octagon inscribed in a circle.
* AF is a diameter of the circle.
* FC is extended to E, such that BCE is a straight line.
* FE is a tangent to the circle at F.
* We need to find the measure of angle FEC.
**2. Properties of a Regular Octagon**
* A regular octagon has 8 equal sides and 8 equal interior angles.
* The measure of each interior angle of a regular octagon is:
* (n - 2) * 180° / n = (8 - 2) * 180° / 8 = 6 * 180° / 8 = 135°
**3. Angles in the Diagram**
* Let O be the center of the circle.
* Since the octagon is regular, the central angle subtended by each side is 360° / 8 = 45°.
* Angle AOF is 45 degrees.
* Angle COF is 45 degrees.
* Angle BOC is 45 degrees.
* Since AF is a diameter, angle ACF is 135/2 = 67.5 degrees, because the angle ACF is half the interior angle of the octagon.
* Angle AFC = 135/2 = 67.5 degrees.
**4. Tangent and Diameter Properties**
* Since FE is a tangent to the circle at F, angle AFE = 90°.
**5. Finding Angle CFE**
* We need to find angle CFE.
* Since angle AFC = 67.5 degrees and angle AFE = 90 degrees.
* angle CFE = angle AFE - angle AFC = 90 - 67.5 = 22.5 degrees.
**6. Finding Angle FEC**
* In triangle FEC, we need to find angle FEC.
* Since BCE is a straight line, angle FCB = 180 - 135 = 45 degrees.
* We know angle CFE = 22.5 degrees.
* In triangle FEC, the sum of the angles is 180 degrees.
* angle FEC + angle CFE + angle FCE = 180 degrees.
* angle FEC + 22.5 + 45 = 180
* angle FEC + 67.5 = 180
* angle FEC = 180 - 67.5
* angle FEC = 112.5 degrees.
**Therefore, the size of angle FEC is 112.5 degrees.**
Question 1170178: 1.a) Deduce that the total surface area S of a cylinder closed at both ends with height, H and base radius R is given by: S=2πR(R+H) where π is a constant.
b) Find the volume of S given that H =15.0m and R =5.0m (leaving your answer in terms of π)
c) Calculate the amount of water the tank can hold leaving your answer in terms of π.
d) Calculate the height (h) of a cuboid tank of cross-sectional area 25 metre cube which has the same capacity as the cylinder tank in (c) above.
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
In your post, in the last question, you write
". . . cross-sectional area 25 metre cube . . . ".
Missis or mister writer, the area is NEVER measures in "metre cube ".
The appropriate unit for the area is " square meter ".
Question 1209832: A body of weight 500N rests on a plane inclined at 20o to the horizontal. The coefficient of friction
is 0.4, determine a force F at an angle of 15o to the plane required to
(a) Pull the body upwards
(b) Push the body downwards
(c) Pull the body downwards
(d) Push the body upwards
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
The solution to this problem in the post by @CPhill is INCORRECT.
The fact is that the applied force at 15° to the inclined plane DECREASES the normal force
produced by the weight.
In the solution by @CPhill the applied force at 15° to the inclined plane, in opposite,
INCREASES the normal force produced by the weight, which is physically absurdist.
Question 1179575: Now explore the position of the redsaw tooth in reference to an imaginary verticalaxis of symmetry of the circular blade. The red toothis initially one foot to the right of the dottedline.How far to the right of this axis is the tooth after37 seconds? After 237 seconds? After t seconds? Drawby hand a graph that shows how the displacement pof the red tooth with respect to the vertical axisis afunction of the elapsed time t.
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
In this post, key necessary input data are missed, due to the author's negligence.
So, the problem is DEFECTIVE: it is mutilated and cannot be solved.
Question 1179574: The graphs of the height h and thehorizontal displacement p of the red saw tooth areexamples of sine and cosine curves, respectively.Draw y = sinx and y = cosx on a graphing tool andcompare them with the graphs that you drew in thepreceding exercises. Use these graphs to answer thefollowing questions:(a)For what values of t is the red tooth 0.8 feetabove the table? 0.8 feet below the table?(b)When is the tooth 6 inches to the right of thevertical axis? When is it farthest left?
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
In this post, key necessary input data are missed, due to the author's negligence.
So, the problem is DEFECTIVE: it is mutilated and cannot be solved.
Question 1179792: Which combination of limit properties is required to evaluate this limit?
lim x-->4 (24/x-2x+2)^3
a. sum, difference, product, root
b. sum, difference, product, power
c. sum, difference, quotient, power
d. sum, difference, quotient, root
e. A limit does not exist.
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
Which combination of limit properties is required to evaluate this limit?
lim x-->4 (24/x-2x+2)^3
a. sum, difference, product, root
b. sum, difference, product, power
c. sum, difference, quotient, power
d. sum, difference, quotient, root
e. A limit does not exist.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
In order for to discuss this problem, we first need to read what is written there.
But i is not possible to read this function properly, since it is written incorrectly.
To write correctly, use parentheses to show clearly which part is the numerator
and which part is the denominator in the formula.
Question 1186907: The world production of gold from 1970 to 1990 can be modelled by G = 5.2t 2 - 76t + 1492, where G is the
number of tonnes of gold and t is the number of years since 1970, t = 1 for 1971 and so on.
a. What was the most amount of gold mined in one year?
b. How much gold was mined in 1978?
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
The world production of gold from 1970 to 1990 can be modelled by G = 5.2t 2 - 76t + 1492,
where G is the number of tonnes of gold and t is the number of years since 1970,
t = 1 for 1971 and so on.
a. What was the most amount of gold mined in one year?
b. How much gold was mined in 1978?
~~~~~~~~~~~~~~~~~~~~~~~~~~~
The problem's formulation is not perfect for a Math problem.
I would edit it in this way (my editing is underlined and crossed)
The world annual production of gold from 1970 to 1990 can be modelled by G = 5.2t 2 - 76t + 1492,
where G is the number of tonnes of gold produced per year and t is the number of years since 1970,
t = 1 for 1971 and so on.
a. What was the  greatest amount of gold mined in one year?
b. How much gold was mined in 1978?
In the post by @CPhill, in the part (a), his reading of the problem is INCORRECT; his interpretation of the problem
is incorrect; his treating of the problem is incorrect; his solution is incorrect and his answer is incorrect.
@CPhill focused on the minimal amount of the gold mined in one year, while the problem in part (a) asks
about the maximum amount.
For it, the amount at the ends of the time interval should be considered.
If you do it, you will find that the greatest amount of the mined gold was achieved in the year 1990 (t = 20),
and this greatest amount was
G(20) = 5.2*20^2 - 76*20 + 1492 = 2052 tons.
Question 1190655: You deposit $200 each month into an account earning 7% interest compounded monthly.
a) How much will you have in the account in 30 years? (Round down to the cents place.)
b) How much total money will you put into the account? (Give the exact value.)
c) How much total interest will you earn?
I am stuck on this; I've tried the formulas for it
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
You deposit $200 each month into an account earning 7% interest compounded monthly.
a) How much will you have in the account in 30 years? (Round down to the cents place.)
b) How much total money will you put into the account? (Give the exact value.)
c) How much total interest will you earn?
I am stuck on this; I've tried the formulas for it
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Calculations in the post by @CPhill are not accurate.
The formula, which should be used is
PV =  .
Calculations should be done without making intermediate rounding.
Rounding is made at the end, ONLY.
The correct answer to (a) is 243,994.20 dollars, which can be rounded to $243,994.
Solved.
------------------------------------------------
I used MS Excel in my computer, which works with 15 decimals after the decimal point,
providing a necessary precision.
It is equivalent to Google worksheet.
These are the best and the most appropriate calculators for such computing.
You write the formula in any text editor, copy-paste it into the spreadsheet cell and get the answer instantly.
Question 1190593: A. Avery and their family currently pay $1300 in rent per month. They want to know how much they could borrow for a mortgage if they wanted to make monthly payments that are the same as their current rent.
1. One possible loan is a 30-year mortgage with a 3.48% APR. How much could Avery borrow? (Round down to the dollar.)
How much would they pay back over the life of the loan?
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
Avery and their family currently pay $1300 in rent per month. They want to know how much they could borrow
for a mortgage if they wanted to make monthly payments that are the same as their current rent.
One possible loan is a 30-year mortgage with a 3.48% APR. How much could Avery borrow? (Round down to the dollar.)
How much would they pay back over the life of the loan?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Calculations in the post by @CPhill are not accurate.
The formula, which should be used, is
PV =  .
Calculations should be done without making intermediate rounding.
Rounding is made at the end, ONLY.
The correct answer is 290,224.56 dollars, which can be rounded to $290,225.
Solved.
------------------------------------------------
I used MS Excel in my computer, which works with 15 decimals after the decimal point,
providing a necessary precision.
It is equivalent to Google worksheet.
These are the best and the most appropriate calculators for such computing.
You write the formula in any text editor, copy-paste it into the spreadsheet cell and get the answer instantly.
Question 1191377: You deposit $400 each month into an account earning 3% interest compounded monthly.
a) How much will you have in the account in 30 years?
b) How much total money will you put into the account?
c) How much total interest will you earn?
Answer by ikleyn(52781) (Show Source):
Question 1191376: How much would you need to deposit in an account each month in order to have $50,000 in the account in 7 years? Assume the account earns 6% interest.
Answer by ikleyn(52781) (Show Source):
Question 1192052: Triplets Peter, Reeta and Nikita have two ways for getting home from school each day: cycle on a tandem bike or walk. The bike can carry either one or two riders at a time. Regardless of the number of people pedalling, cycling speed is 5 times walking speed. The triplets always leave school at the same time and always use the same path between school and home, whether walking or cycling. The school is 5 km from home and their walking speed is 4 kilometres per hour.
a) On Monday, Nikita and Peter cycle and Reeta walks. On reaching the point four-fifths of the way home the bike gets a puncture, so Nikita and Peter walk the rest of the way home. How far from school is Peeta when the cyclists arrive home?
b) On Tuesday, Peter and Reeta ride the bike and Nikita walks. When the cyclists arrive home, Peter hops off the bike and Reeta rides back towards school to collect Nikita. How far from school is Nikita when Reeta reaches her?
c) On Wednesday, Reeta and Nikita take the bike and Peter walks. When the cyclists are halfway home, Reeta hops off and walks the rest of the way, while Nikita heads back to pick up Peter. How far from school is Reeta when her siblings pass her on the bike?
d) On Thursday, it is Reeta's turn to walk. Peter drops Nikita off at a certain point leaving her to walk home. Meanwhile he returns to pick up Reeta and they cycle home together. If all three arrive home at the same time, how far from school are the drop-off and pick-up points?
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
Triplets Peter, Reeta and Nikita have two ways for getting home from school each day: cycle on a tandem bike or walk.
The bike can carry either one or two riders at a time.
Regardless of the number of people pedalling, cycling speed is 5 times walking speed.
The triplets always leave school at the same time and always use the same path between school and home,
whether walking or cycling. The school is 5 km from home and their walking speed is 4 kilometres per hour.
a) On Monday, Nikita and Peter cycle and Reeta walks. On reaching the point four-fifths of the way home
the bike gets a puncture, so Nikita and Peter walk the rest of the way home. How far from school is Peeta when the cyclists arrive home?
b) On Tuesday, Peter and Reeta ride the bike and Nikita walks. When the cyclists arrive home,
Peter hops off the bike and Reeta rides back towards school to collect Nikita. How far from school is Nikita when Reeta reaches her?
c) On Wednesday, Reeta and Nikita take the bike and Peter walks. When the cyclists are halfway home,
Reeta hops off and walks the rest of the way, while Nikita heads back to pick up Peter.
How far from school is Reeta when her siblings pass her on the bike?
d) On Thursday, it is Reeta's turn to walk. Peter drops Nikita off at a certain point leaving her to walk home.
Meanwhile he returns to pick up Reeta and they cycle home together. If all three arrive home at the same time,
how far from school are the drop-off and pick-up points?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
In the solution by @CPhill to this problem, his interpretation for part (d) is CERTAINLY incorrect.
The correct interpretation for part (d), from the problem, is that first Peter and Nikita are cycling,
while Reeta is  - - - not waiting at the school.
Then Peter drops Nikita off at a certain point leaving her to walk home, and Peter returns back
to pick up Reeta on her way to the home.
After that, Peter and Reeta are cycling home together.
It is the ONLY one and the UNIQUE possible interpretation, and the text points it directly, saying in (d) that
" On Thursday, it is Reeta's turn to walk ".
It is a classic Travel and Distance problem, which I know along and across from my school years.
So, below is my solution for part (d) in this correct interpretation.
So, the distance from the school to home is 5 km; walking speed is 4 km/h and the cycling speed is 20 km/h.
Let x be the distance from the school to the drop-off point and
let y be the distance from the school to the pick-up point.
The time for Peter to get the drop-off point is  hours.
The time for Peter to move from the drop-off point to pick-up point is  hours.
The time for Peter to get the drop-off point and then to return to the pick-up point is the sum
 =  . (1)
This time, hours, is equal to the time for Reeta to get the pick-up pont from the school, which is  hours.
It gives us this equation
= . (2)
Simplify it
4(2x-y) = 20y, or 2x-y = 5y, or 2x = 5y + y, or 2x = 6y, or x = 3y. (3).
+------------------------------------------------------------------------+
| At this point, we complete our consideration for moving Peter |
| from the school to the drop-off point and then to the pick-up point. |
+------------------------------------------------------------------------+
Now we will consider the other part of moving.
The time for Nikita to get the home from the drop-off point walking is  hours.
During this time, Peter is cycling from point x to point y and then from point y to the home.
It takes for Peter +  =  hours.
The time for Nikita is the same as the time for Peter.
It gives us this equation
= . (4)
Simplify it
20(5-x) = 4(5+x- 2y), or 100-20x = 20+4x-8y, or 100-20 = 4x + 20x-8y, or 24x-8y = 80, or 3x - y = 10. (5)
Thus, we have now two equations
x = 3y and 3x - y = 10.
Substitute first equation, x = 3y, into the second equation
3*(3y) - y = 10, or 9y - y = 10, 8y = 10, y =  =  = 1.25.
Then x = 3y = 3*1.25 = 3.75.
Thus the drop-off point is 3.75 km from the school, and the pick-up point is 1.25 km from the school. ANSWER
At this point, the part (d) of the problem is solved completely.
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
Regarding the post by @CPhill . . .
Keep in mind that @CPhill is a pseudonym for the Google artificial intelligence.
The artificial intelligence is like a baby now. It is in the experimental stage
of development and can make mistakes and produce nonsense without any embarrassment.
It has no feeling of shame - it is shameless.
This time, again, it made an error.
Although the @CPhill' solution are copy-paste Google AI solutions, there is one essential difference.
Every time, Google AI makes a note at the end of its solutions that Google AI is experimental
and can make errors/mistakes.
All @CPhill' solutions are copy-paste of Google AI solutions, with one difference:
@PChill never makes this notice and never says that his solutions are copy-past that of Google.
So, he NEVER SAYS TRUTH.
Every time, @CPhill embarrassed to tell the truth.
But I am not embarrassing to tell the truth, as it is my duty at this forum.
And the last my comment.
When you obtain such posts from @CPhill, remember, that NOBODY is responsible for their correctness,
until the specialists and experts will check and confirm their correctness.
Without it, their reliability is ZERO and their creadability is ZERO, too.
Question 1164837: Mark left town P for town Q at 10am, travelling at a uniform speed of 75km/h. Jane left town P for town Q an hour after Mark, travelling at 120km/h. She passed Mark after travelling 2/5 of the journey. she reached town Q at 4 pm. How far was Jane from town Q when she passed Mark??
Found 3 solutions by MathTherapy, greenestamps, ikleyn:
Answer by MathTherapy(10552)  (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The given information is inconsistent, so the problem can't be solved.
Jane left at 11 am and arrived at 4 pm, traveling 5 hours at a speed of 120 km/hr, so the distance between the two towns was 120*5 = 600 km.
She passed Mark after traveling 2/5 of the way. 2/5 of 5 hours is 2 hours; 2/5 of 600 km is 240 km. So she passed Mark after traveling 240 km.
But Mark left 1 hour before Jane, so he traveled for 2+1 = 3 hours at 75 km/hr, traveling a distance of 3*75 = 225 km, by the time that Jane passed him.
And it is clearly impossible for Jane to have passed Mark when she 240 km and he was 225 km from their common starting point.
ANSWER: bad information; no solution
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
Mark left town P for town Q at 10am, travelling at a uniform speed of 75km/h.
Jane left town P for town Q an hour after Mark, travelling at 120km/h.
She passed Mark after travelling 2/5 of the journey. she reached town Q at 4 pm.
How far was Jane from town Q when she passed Mark?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
When Jane left town P for town Q at 11 am, Mark was 75 km ahead.
The difference of their speeds is 120 - 75 = 45 km/h.
Hence, Jane passed Mark in  =  hours, counting from 11 am.
During this time, hours, Jane covered the distance of  = 40*5 = 200 km.
This distance, 200 km, is  of the distance from P to Q, according to the problem.
Hence, the whole distance from P to Q is  = 500 km.
From the other side, Jane traveled 5 hours (from 11 am to 4 pm) at the speed of 120 km/h -
hence, she traveled 5*120 = 600 km.
Thus we obtained two different values, 500 km and 600 km, for the distance from P to Q.
It tells us that the different parts of the problem are not consistent.
Thus, the problem is SELF-CONTRADICTORY, is posed INCORRECTLY and MAKES NO sense.
ANSWER. As presented in the post, the problem is SELF-CONTRADICTORY,
is posed INCORRECTLY and MAKES NO sense.
The problem in the post is composed in absolutely illiterate way.
All the accusations should be addressed to the problem's creator.
Question 1165007: The Cream and Custard Bakery makes both coffee cakes and Danish in large pans. The main
ingredients are flour and sugar. There are 25 pounds of flour and 16 pounds of sugar available
and the maximum demand for coffee cakes is 8. Five pounds of flour and 2 pounds of sugar
are required to make one pan of coffee cake, and 5 pounds of flour and 4 pounds of sugar are
required to make one pan of Danish. One pan of coffee cake has a profit of PhP 1, and one pan
of Danish has a profit of PhP 5. Determine the number of pans of cake and Danish that the
bakery must produce each day so that profit will be maximized.
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
The Cream and Custard Bakery makes both coffee cakes and Danish in large pans. The main
ingredients are flour and sugar. There are 25 pounds of flour and 16 pounds of sugar available
and the maximum demand for coffee cakes is 8. Five pounds of flour and 2 pounds of sugar
are required to make one pan of coffee cake, and 5 pounds of flour and 4 pounds of sugar are
required to make one pan of Danish. One pan of coffee cake has a profit of PhP 1, and one pan
of Danish has a profit of PhP 5. Determine the number of pans of cake and Danish that the
bakery must produce each day so that profit will be maximized.
~~~~~~~~~~~~~~~~~~~~~~~~~~~
This problem is simple: it is accessible to a third-grader.
Therefore, I will not write a standard solution using Linear programming method,
but will show a simple arithmetic solution.
From the problem, it is clear that the maximum number of cakes is 5.
Indeed, the bakery can make 5 coffee cakes, using 5*5 = 25 pounds of flour and 5*2 = 10 pounds of sugar.
It can not make more than 5 cakes, either coffee cakes or Danish.
So, our task is to make a schedule (a Table) showing possible cakes that can be prepared, and possible profit.
Obviously, we should spend as much of the ingredients as allowed by restrictions.
Also, from the given data, it is obvious, that for any given number of total cakes, the bakery should
make as many Danish pans as possible, then adding coffee cakes until fitting the restrictions.
It is because the first limiting restriction is the amount of floor, and regarding the floor,
the coffee cakes and the Danish pans are in equal position, while Danish provides greater profit.
the number coffee Danish Floor Sugar Profit
of cakes cakes pans (Php)
(total)
--------------------------------------------------------------------------
5 5 0 5*5 = 25 5*2 = 10 5*1 = 5
4 0 4 0*5+4*5 = 20 0*2+4*4 = 16 4*5 = 20 (*)
4 1 3 1*5+3*5 = 20 1+3*5 = 16
4 2 2 2*5+2*5 = 20 2*2+2*4 = 12 2*2+2*4 = 12
There is no sense to continue the table further.
Looking into the table, we mark the optimum solution by (*).
It provides the maximum profit of Php 20 .
ANSWER. 4 Danish pans and 0 (zero) coffee cakes satisfy the restrictions and provide the maximum profit of Php 20.
Solved.
Question 1167347: You started a business making sculptures that sell for 14.50 each. It cost you $.75 to make each sculpture. It takes you two hours to make a sculpture. You receive 50% off the selling price from the store selling the sculptures. She spends 14 hours a week making sculptures how many weeks will it take you to earn a profit of $360
Answer by ikleyn(52781) (Show Source):
Question 1167600: for each of the following, compute the function value for the real given number t by hand (without a calculator)
tan (t), where t is the arc of length 2pi/3 oriented clockwise on the unit circle
cot(s), where s is the arc with terminal point(-1,0)
sin(t), where t = 17pi/6
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
D U P L I C A T E
See the solution under this link
https://www.algebra.com/algebra/homework/Finance/Finance.faq.question.1167605.html
Question 1167605: for each of the following, compute the function value for the real given number t by hand (without a calculator)
tan (t), where t is the arc of length 2pi/3 oriented clockwise on the unit circle
cot(s), where s is the arc with terminal point(-1,0)
sin(t), where t = 17pi/6
Answer by ikleyn(52781) (Show Source):
Question 1210258: Given
Principal $10,000, Interest Rate 8%, Time 240 days (use ordinary interest)
Partial payments: On 100th day, $4,000; On 180th day, $2,000
Use the U.S. Rule to solve for total interest cost.
Note: Use 360 days a year. Do not round intermediate calculations. Round your answer to the nearest cent.
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
This post makes no sense.
As a mathematical problem, it is INCOMPLETE, and, therefore, can not be solved.
The question in the post is missed.
The right place for this composition is a garbage bin.
Question 1166360: As a receptionist for a hospital, one of Elizabeth's tasks is to schedule appointments. She allots 60 minutes for the first visit and 30 minutes for a follow-up. The doctor cannot perform more than eight follow-ups per day. The hospital has eight hours (480 minutes) available for appointments. The first visit costs $120 and the follow-up costs $70. Let x be the number of first visits and y be the number of follow-ups. Write a system of inequalities to represent the number of first visits and the number of follow-ups that can be performed.
What is the maximum income that the doctor receives per day?
a. $960 b. $1,040
c. $970 d. $1,920
Determine the number of first visits and follow-ups to be scheduled to make the maximum income.
a. 16 first visits and 0 follow-ups b. 4 first visits and 7 follow-ups
c. 8 first visits and 0 follow-ups d. 4 first visits and 8 follow-ups
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
As a receptionist for a hospital, one of Elizabeth's tasks is to schedule appointments.
She allots 60 minutes for the first visit and 30 minutes for a follow-up.
The doctor cannot perform more than eight follow-ups per day.
The hospital has eight hours (480 minutes) available for appointments.
The first visit costs $120 and the follow-up costs $70.
Let x be the number of first visits and y be the number of follow-ups.
Write a system of inequalities to represent the number of first visits and the number of follow-ups
that can be performed.
What is the maximum income that the doctor receives per day?
a. $960 b. $1,040
c. $970 d. $1,920
Determine the number of first visits and follow-ups to be scheduled to make the maximum income.
a. 16 first visits and 0 follow-ups b. 4 first visits and 7 follow-ups
c. 8 first visits and 0 follow-ups d. 4 first visits and 8 follow-ups
~~~~~~~~~~~~~~~~~~~~~~~~~~~~
If the goal is to solve the minimax problem and to find the maximum income (with learning something useful on the way),
then the problem can be solved MENTALLY without writing any equations and/or any inequalities,
and without making trial-and-error calculations.
From the problem, two follow-ups visits require 30+30 = 60 minutes and provide the income
of $70+$70 = $140. It is greater than one first visit, which requires the same time of 60 minutes
and provides the income of $120.
Therefore, the idea is to schedule as many pairs of 30 minutes follow-ups visits as it is possible
under the problem's restrictions, and then to fill the rest of the time by 60-minutes first visits.
Since the doctor has the restriction to serve no more than 8 follow-ups visitors per day,
it IDEALLY fits to this scheme:
8 follow-ups visits PLUS 4 first time visits,
giving the maximum possible income of 8*70 + 4*120 = 1040 dollars per day and satisfying all restrictions.
Solved.
In my view, this way solving/reasoning is MUCH better than to follow the instructions in the post.
Why ? - - - Because it teaches to think and to see the solution via thinking, instead of
simply follow instructions and making trials-and-errors, as if a person/(a student) is mentally blind.
Question 1168239: 1. A manufacturer sells kayaks to dealers for Php 94,500 with discounts of 40/10/5.
a. Determine the net price of a kayak.
b. How much is the total trade discount?
c. What is the equivalent single trade discount for this transaction?
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let the list price of a kayak be $L = \text{Php } 94,500$.
The discounts offered are 40/10/5. This means there are three successive discounts.
**a. Determine the net price of a kayak.**
To find the net price, we apply the discounts sequentially to the original price.
1. **First discount (40%):**
Discount amount = $40\% \times 94,500 = 0.40 \times 94,500 = \text{Php } 37,800$
Price after the first discount = $94,500 - 37,800 = \text{Php } 56,700$
2. **Second discount (10%):**
This discount is applied to the price after the first discount.
Discount amount = $10\% \times 56,700 = 0.10 \times 56,700 = \text{Php } 5,670$
Price after the second discount = $56,700 - 5,670 = \text{Php } 51,030$
3. **Third discount (5%):**
This discount is applied to the price after the second discount.
Discount amount = $5\% \times 51,030 = 0.05 \times 51,030 = \text{Php } 2,551.50$
Net price = $51,030 - 2,551.50 = \text{Php } 48,478.50$
Alternatively, we can calculate the net price using the discount factors:
Net price = List price $\times (1 - \text{discount}_1) \times (1 - \text{discount}_2) \times (1 - \text{discount}_3)$
Net price = $94,500 \times (1 - 0.40) \times (1 - 0.10) \times (1 - 0.05)$
Net price = $94,500 \times (0.60) \times (0.90) \times (0.95)$
Net price = $94,500 \times 0.513$
Net price = $\text{Php } 48,478.50$
**b. How much is the total trade discount?**
The total trade discount is the difference between the original list price and the net price.
Total trade discount = List price - Net price
Total trade discount = $94,500 - 48,478.50 = \text{Php } 46,021.50$
**c. What is the equivalent single trade discount for this transaction?**
The equivalent single trade discount is the single discount rate that would give the same net price. Let the equivalent single discount rate be $d$.
Net price = List price $\times (1 - d)$
$48,478.50 = 94,500 \times (1 - d)$
$1 - d = \frac{48,478.50}{94,500}$
$1 - d = 0.513$
$d = 1 - 0.513$
$d = 0.487$
To express this as a percentage, we multiply by 100:
Equivalent single trade discount = $0.487 \times 100\% = 48.7\%$
Final Answer: The final answer is $\boxed{a. \text{Php } 48,478.50, b. \text{Php } 46,021.50, c. 48.7\%}$
Question 1210206:
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
The post came with no question, so I deleted it, for your education.
Each post, submitted to this forum, must be complete, including a question.
For the rules regarding posting to this forum, see this web-page
https://www.algebra.com/tutors/students/ask.mpl?action=ask_question&topic=Equations&return_url=http://www.algebra.com/algebra/homework/equations/
from which you post your problems.
It is assumed that you read these rules before posting.
It is also assumed that you do understand what is written in that page and follow the rules.
Those who violate the rules, work against their own interests.
Question 1168949: A quilt is made up of strips of
cloth, starting with an inner square surrounded
by rectangles to form successively larger
squares. The inner square and all rectangles
have a width of 1 foot. Write an expression
using summation notation that gives the sum of
the areas of all the strips of cloth used to make
the quilt shown. Then evaluate the expression.
Found 2 solutions by ikleyn, CPhill:
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
A quilt is made up of strips of cloth, starting with an inner square
surrounded by rectangles to form successively larger squares.
The inner square and all rectangles have a width of 1 foot.
Write an expression using summation notation that gives the sum of
the areas of all the strips of cloth used to make the quilt shown.
Then evaluate the expression.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Step 1. At step 1, we have the inner 1x1-square in the center.
It is surrounded by 4 (four) rectangles of the length 2 ft and the width 1 ft.
These 5 shapes form a 3x3-square with the area 3*3 = 9 square feet.
At this point, we can write this equality
1 + 4*2 = 9 square feet.
Step 2. At step 2, we have this 3x3-square surrounded by 4 (four) rectangles
of the length 4 ft and the width 1 ft.
Altogether, they form 5x5-square with the area 5*5 = 25 square feet.
At this point, we can write this equality
1 + 4*2 + 4*4 = 25 square feet.
Step 3. At step 3, we have this 5x5-square surrounded by 4 (four) rectangles
of the length 6 ft and the width 1 ft.
Altogether, they form 7x7-square with the area 7*7 = 49 square feet.
At this point, we can write this equality
1 + 4*2 + 4*4 + 4*6 = 49 square feet.
Step 4. At step 4, we have this 7x7-square surrounded by 4 (four) rectangles
of the length 8 ft and the width 1 ft.
Altogether, they form 9x9-square with the area 9*9 = 81 square feet.
At this point, we can write this equality
1 + 4*2 + 4*4 + 4*6 + 4*8 = 81 square feet.
. . . . . and so on . . . . .
The pattern is just seen. For clarity, I will describe the common step 'n'.
Step n. At step n, we have (2n-1) x (2n-1)-square from the previous step,
surrounded by 4 (four) rectangles of the length 2n ft and the width 1 ft.
Altogether, they form (2n+1) x (2n+1)-square with the area (2n+1)*(2n+1) = (2n+1)^2 square feet.
At this point, we can write this equality
1 + 4*2 + 4*4 + 4*6 + 4*8 + . . . + 4(2n) = (2n+1)^2.
It can be proved by the method of Mathematical induction.
The step of induction is this
Prove that (2n-1)^2 + 4*(2n) = (2n+1)^2.
To prove it, simply open parentheses in the left side and simplify
(2n-1)^2 + 4*(2n) = 4n^2 - 2*(2n) + 1 + 4*(2n) = 4n^2 + 2*(2n) + 1 = (2n+1)^2.
So, the proof is in one line.
It is obvious at the same degree, as the geometric step-by-step procedure described above in my post.
At this point, all explanations are complete and the problem is solved in full.
--------------------
Tutor @HPhill (which represents and uses an Artificial Intelligence)
did not get understanding the problem and shot out the target to NOWHERE.
Simply ignore his post. This Artificial Intelligence is still undertrained.
It works perfectly, it it finds a source in its database to re-write from,
but can not think independently.
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's analyze the pattern of the quilt construction:
* **Strip 1 (Inner Square):** This is a square with a width of 1 foot. Since it's a square, its length is also 1 foot.
* Area of Strip 1 = $1 \times 1 = 1$ square foot.
* **Strip 2 (First set of rectangles):** This strip surrounds the inner square. It consists of four rectangles, each with a width of 1 foot. The length of each rectangle is equal to the side length of the inner square (1 foot).
* Area of each rectangle = $1 \times 1 = 1$ square foot.
* Total area of Strip 2 = $4 \times 1 = 4$ square feet.
* **Strip 3 (Second set of rectangles):** This strip surrounds the shape formed by the inner square and the first strip of rectangles. The side length of this new square is $1 + 1 + 1 = 3$ feet. This strip also consists of four rectangles with a width of 1 foot. The length of each rectangle is equal to the side length of the square it's being added to (3 feet).
* Area of each rectangle = $1 \times 3 = 3$ square feet.
* Total area of Strip 3 = $4 \times 3 = 12$ square feet.
* **Strip 4 (Third set of rectangles):** This strip surrounds the shape formed by the previous strips. The side length of this new square is $3 + 1 + 1 = 5$ feet. This strip consists of four rectangles with a width of 1 foot and a length equal to the side length of the square it's being added to (5 feet).
* Area of each rectangle = $1 \times 5 = 5$ square feet.
* Total area of Strip 4 = $4 \times 5 = 20$ square feet.
We can see a pattern here. Let $n$ be the strip number (starting with the inner square as $n=1$). The side length of the square just before adding the $n$-th strip can be expressed as $2(n-1) - 1 + 2 = 2n - 2 + 1 = 2n - 1$ for $n > 1$, and 1 for $n=1$. However, it's easier to see the sequence of side lengths of the squares: 1, 3, 5, ... which is $2n - 1$.
The length of the rectangles in the $n$-th strip (for $n > 1$) is the side length of the square formed by the first $n-1$ strips, which is $2(n-1) - 1 = 2n - 3$. The width is always 1. There are 4 such rectangles. So the area of the $n$-th strip (for $n > 1$) is $4(2n - 3)$.
Let's re-examine the pattern of the total area of each strip:
Strip 1: 1
Strip 2: 4
Strip 3: 12
Strip 4: 20
This doesn't directly fit $4(2n-3)$. Let's think about the increase in area at each step. When we add a strip of width 1 around a square of side length $s$, we add four rectangles of size $1 \times s$.
Let $s_n$ be the side length of the square after adding the $n$-th strip.
$s_0 = 1$ (inner square)
$s_1 = 3$
$s_2 = 5$
$s_n = 2n + 1$ (if we consider the number of added strips)
Let's consider the strips.
Strip 1 (n=1): Area = $1^2 = 1$.
Strip 2 (n=2): Adds around a $1 \times 1$ square. Four rectangles of $1 \times 1$. Area = $4 \times 1 = 4$.
Strip 3 (n=3): Adds around a $3 \times 3$ square. Four rectangles of $1 \times 3$. Area = $4 \times 3 = 12$.
Strip 4 (n=4): Adds around a $5 \times 5$ square. Four rectangles of $1 \times 5$. Area = $4 \times 5 = 20$.
Strip $n$ (for $n > 1$): Adds around a $(2n-3) \times (2n-3)$ square. Four rectangles of $1 \times (2n-3)$. Area = $4(2n-3)$.
Let $A_i$ be the area of the $i$-th strip.
$A_1 = 1$
$A_i = 4(2i - 3)$ for $i > 1$.
Suppose the quilt shown has $k$ strips. The sum of the areas of all the strips is:
Sum $= A_1 + \sum_{i=2}^{k} A_i = 1 + \sum_{i=2}^{k} 4(2i - 3)$
Let's assume the quilt shown has 4 strips (based on our initial analysis).
Sum $= 1 + \sum_{i=2}^{4} 4(2i - 3) = 1 + [4(2(2) - 3) + 4(2(3) - 3) + 4(2(4) - 3)]$
Sum $= 1 + [4(4 - 3) + 4(6 - 3) + 4(8 - 3)]$
Sum $= 1 + [4(1) + 4(3) + 4(5)]$
Sum $= 1 + [4 + 12 + 20] = 1 + 36 = 37$ square feet.
Now let's write the expression using summation notation for a quilt with $k$ strips:
Sum $= 1 + \sum_{i=2}^{k} 4(2i - 3)$
To evaluate this expression for $k=4$:
Sum $= 1 + 4 \sum_{i=2}^{4} (2i - 3) = 1 + 4 [(2(2) - 3) + (2(3) - 3) + (2(4) - 3)]$
Sum $= 1 + 4 [(4 - 3) + (6 - 3) + (8 - 3)]$
Sum $= 1 + 4 [1 + 3 + 5] = 1 + 4 [9] = 1 + 36 = 37$.
Final Answer: The final answer is $\boxed{\text{Expression: } 1 + \sum_{i=2}^{k} 4(2i - 3), \text{ Evaluation (for 4 strips): } 37}$
Question 1168762: 1. sketch 2 periods of the graph of the function f(x) = tan(2x)
2.sketch one period of the of the graphs of y =cos(x) and f(x) = -2cos(x/2) on the same coordinate plane. Include all five key in your graph
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
Go to website www.desmos.com/calculator
Find there free of charge online plotting tool for common use.
Print equations of your functions there and get the graphs instantly.
Question 1167970: Deshaun the trainer has two solo workout plans that he offers his clients: Plan A and Plan B. Each client does either one or the other (not both). On Wednesday there were 8 clients who did Plan A and 4 who did Plan B. On Thursday there were 3 clients who did Plan A and 2 who did Plan B. Deshaun trained his Wednesday clients for a total of 9 hours and his Thursday clients for a total of 4 hours. How long does each of the workout plans last?
Answer by ikleyn(52781) (Show Source):
|
Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790, 2791..2835, 2836..2880, 2881..2925, 2926..2970, 2971..3015, 3016..3060, 3061..3105, 3106..3150, 3151..3195, 3196..3240, 3241..3285, 3286..3330, 3331..3375, 3376..3420, 3421..3465, 3466..3510, 3511..3555, 3556..3600, 3601..3645, 3646..3690, 3691..3735, 3736..3780, 3781..3825, 3826..3870, 3871..3915, 3916..3960, 3961..4005, 4006..4050, 4051..4095, 4096..4140, 4141..4185, 4186..4230, 4231..4275, 4276..4320, 4321..4365, 4366..4410, 4411..4455, 4456..4500, 4501..4545, 4546..4590, 4591..4635, 4636..4680, 4681..4725, 4726..4770, 4771..4815, 4816..4860, 4861..4905, 4906..4950, 4951..4995, 4996..5040, 5041..5085, 5086..5130, 5131..5175, 5176..5220, 5221..5265, 5266..5310, 5311..5355, 5356..5400, 5401..5445, 5446..5490, 5491..5535, 5536..5580, 5581..5625, 5626..5670, 5671..5715, 5716..5760, 5761..5805, 5806..5850, 5851..5895, 5896..5940, 5941..5985, 5986..6030, 6031..6075, 6076..6120, 6121..6165, 6166..6210, 6211..6255, 6256..6300, 6301..6345, 6346..6390, 6391..6435, 6436..6480, 6481..6525, 6526..6570, 6571..6615, 6616..6660, 6661..6705, 6706..6750, 6751..6795, 6796..6840, 6841..6885, 6886..6930, 6931..6975, 6976..7020, 7021..7065, 7066..7110, 7111..7155, 7156..7200, 7201..7245, 7246..7290, 7291..7335, 7336..7380, 7381..7425, 7426..7470, 7471..7515, 7516..7560, 7561..7605, 7606..7650, 7651..7695, 7696..7740, 7741..7785, 7786..7830, 7831..7875, 7876..7920, 7921..7965, 7966..8010, 8011..8055, 8056..8100, 8101..8145, 8146..8190, 8191..8235, 8236..8280, 8281..8325, 8326..8370, 8371..8415, 8416..8460, 8461..8505, 8506..8550, 8551..8595, 8596..8640, 8641..8685, 8686..8730, 8731..8775, 8776..8820, 8821..8865, 8866..8910, 8911..8955, 8956..9000, 9001..9045, 9046..9090, 9091..9135, 9136..9180, 9181..9225, 9226..9270, 9271..9315, 9316..9360, 9361..9405, 9406..9450, 9451..9495, 9496..9540, 9541..9585, 9586..9630, 9631..9675, 9676..9720, 9721..9765, 9766..9810, 9811..9855, 9856..9900, 9901..9945, 9946..9990, 9991..10035, 10036..10080, 10081..10125, 10126..10170, 10171..10215, 10216..10260, 10261..10305, 10306..10350, 10351..10395, 10396..10440, 10441..10485, 10486..10530, 10531..10575, 10576..10620, 10621..10665, 10666..10710, 10711..10755, 10756..10800, 10801..10845, 10846..10890, 10891..10935, 10936..10980, 10981..11025, 11026..11070, 11071..11115, 11116..11160, 11161..11205, 11206..11250, 11251..11295, 11296..11340, 11341..11385, 11386..11430, 11431..11475, 11476..11520, 11521..11565, 11566..11610, 11611..11655, 11656..11700, 11701..11745, 11746..11790, 11791..11835, 11836..11880, 11881..11925, 11926..11970, 11971..12015, 12016..12060, 12061..12105, 12106..12150, 12151..12195, 12196..12240, 12241..12285, 12286..12330, 12331..12375, 12376..12420, 12421..12465, 12466..12510, 12511..12555, 12556..12600, 12601..12645, 12646..12690, 12691..12735, 12736..12780, 12781..12825, 12826..12870, 12871..12915, 12916..12960, 12961..13005, 13006..13050, 13051..13095, 13096..13140, 13141..13185, 13186..13230, 13231..13275, 13276..13320, 13321..13365, 13366..13410, 13411..13455, 13456..13500, 13501..13545, 13546..13590, 13591..13635, 13636..13680, 13681..13725, 13726..13770, 13771..13815, 13816..13860, 13861..13905, 13906..13950, 13951..13995, 13996..14040, 14041..14085, 14086..14130, 14131..14175, 14176..14220, 14221..14265, 14266..14310, 14311..14355, 14356..14400, 14401..14445, 14446..14490, 14491..14535, 14536..14580, 14581..14625, 14626..14670, 14671..14715, 14716..14760, 14761..14805, 14806..14850, 14851..14895, 14896..14940, 14941..14985, 14986..15030, 15031..15075, 15076..15120, 15121..15165, 15166..15210, 15211..15255, 15256..15300, 15301..15345, 15346..15390, 15391..15435, 15436..15480, 15481..15525, 15526..15570, 15571..15615, 15616..15660, 15661..15705, 15706..15750, 15751..15795, 15796..15840, 15841..15885, 15886..15930, 15931..15975, 15976..16020, 16021..16065, 16066..16110, 16111..16155, 16156..16200, 16201..16245, 16246..16290, 16291..16335, 16336..16380, 16381..16425, 16426..16470, 16471..16515, 16516..16560
|
