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all answered: 182 problems
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Equations/1206049: A woman wishes to enclose a rectangular garden with fencing, using the side of her garage as one side of the rectangle. A neighbor gave her 33 feet of fencing, and she wants the length of the garden along the garage to be 9 feet more than the width. What are the dimensions of the garden?
The length of the rectangular garden is ? and the width of the rectangular garden is ?
1 solutions
Answer 843236 by josgarithmetic(39158)   on 2024-02-11 21:57:26 (Show Source):
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Pythagorean-theorem/1206046: Abigail's bedroom is rectangular. The length of one wall of Abigail's bedroom is 6 meters. The length from one corner of the bedroom to the diagonally opposite corner is 7 meters. What is the length of the other wall? If necessary, round to the nearest tenth.
1 solutions
Answer 843235 by Edwin McCravy(19626)  on 2024-02-11 19:21:47 (Show Source):
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Miscellaneous_Word_Problems/1206044: Mr. Smith had just finished baking a cake for the mathematicians’ banquet. The cake was specially designed in the shape of a cube. In the process of carrying the cake to the frosting table, Mr. Smith suddenly slipped and the cube-cake went sailing into the vat of chocolate frosting. Mr. Smith thought quickly and then yelled “FIRE!” (He knew that no one would come to help if he yelled, “Chocolate”).
Almost immediately help arrived and the cake was fished out of the chocolate. Fortunately, the cube-cake was still in one piece, but was now frosted on all sides.
Mr. Smith proceeded to the banquet hall with his unusually frosted cake in hand.
The mathematicians were delighted when they saw the cube-cake with all of the frosting. They asked Mr. Smith to stay and cut the cake.
One of the mathematicians, Mrs. Hayne suggested that the cake be cut into cube-shaped pieces, all pieces the same size. Mr. Smith agreed, but before cutting the cake he turned to Mrs. Hayne and asked how many mathematicians would like a piece without frosting, a piece with only one side frosted, a piece with exactly two sides frosted, or a piece with three sides frosted.
Being a mathematician, Mrs. Hayne responded, “Cut the cake so that the number of pieces without frosting is equal to eight times the number of pieces that have frosting on three sides. You will then have enough of each type of piece to satisfy everyone, with nothing left over.”
Please answer the following questions.
How many mathematicians attended the banquet?
How many mathematicians were served a piece of cake without frosting? How many mathematicians requested a piece with only one side frosted? How many were served a piece of cake with exactly two sides frosted? How many requested a piece with three sides frosted?
1 solutions
Answer 843229 by math_tutor2020(3057) on 2024-02-11 13:57:17 (Show Source):
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Complex_Numbers/1206040: Without exp anding , prove that the following det er min atnts vanish ., |{{ b - c , c - a , c },{ c , b - x , a },{ c , a , c - x }}| = zero
1 solutions
Answer 843222 by math_tutor2020(3057) on 2024-02-11 11:23:44 (Show Source):
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Probability-and-statistics/1206038: size : 10,12,14,16,18,20
frequency: 3,7,?,20,8,5
mean is 15 find the missing frequency
1 solutions
Answer 843221 by math_tutor2020(3057) on 2024-02-11 10:52:03 (Show Source):
You can put this solution on YOUR website!
x = the missing frequency value
Given data
| Size | 10 | 12 | 14 | 16 | 18 | 20 | | Frequency | 3 | 7 | x | 20 | 8 | 5 |
Or we can format the table like this
| Size | Frequency | | 10 | 3 | | 12 | 7 | | 14 | x | | 16 | 20 | | 18 | 8 | | 20 | 5 |
Multiply each size with its paired frequency. Then add up the results.
10*3 + 12*7 + 14*x + 16*20 + 18*8 + 20*5 = 14x+678
This is then divided over the total frequency (3+7+x+20+8+5 = x+43) to determine the mean.
mean = (14x+678)/(x+43) = 15
(14x+678)/(x+43) = 15
14x+678 = 15(x+43)
14x+678 = 15x+645
678-645 = 15x-14x
33 = x
x = 33
Answer: 33
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Miscellaneous_Word_Problems/1205918: 2000 shares of stocks and bonds sold for $200,000. Bonds were sold for $400 each. Preferred stock sold at 2 for $100. Common stock (employees only) sold at 4 for $100. Determine number of common stock sold.
Not sure how to solve.
2 solutions
Answer 843209 by MathTherapy(10418)  on 2024-02-10 16:38:41 (Show Source):
You can put this solution on YOUR website!
2000 shares of stocks and bonds sold for $200,000. Bonds were sold for $400 each. Preferred stock sold at 2 for $100. Common stock (employees only) sold at 4 for $100. Determine number of common stock sold.
Not sure how to solve.
I found that the least number of bonds sold would've been greater than 285, and the greatest number would've been
400. I also found that more than 400 (401) bonds could not have been sold, as that number would've resulted in 1,614
shares of common stock and - 15 shares of preferred stock. In addition, the sale of 285 bonds would also yield
nonsensical results. So, the number of bonds that could've been sold would've been between 286 and 400, inclusive, a
TOTAL of 115 different possibilities. However, because bonds come in $50 increments/denominations, the least and
greatest amount that could've been sold would've been 300 and 400, respectively.
So, we now have 3 $50-increment possibilities for bonds, from 300 - 400, as follows:
----------------------------------------------
|Instrument | No. | No. | No. |
|BONDS | 300| 350| 400|
|Shares of COMMON STOCK | 200|1,250| 0|
|Shares of PREFERRED STOCK |1,500| 750|1,600|
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|Total |2,000|2,000|2,000|
----------------------------------------------
Now, since 400 bonds lead to 0 common stocks being sold - and it's stated that common and preferred were sold - I
would rule out the 400-bond, 1,600 preferred-stock sale.
This now leaves the following 2 possibilities:
----------------------------------------
|Instrument | No. | No. |
|BONDS | 300| 350|
|Shares of COMMON STOCK | 200|1,250|
|Shares of PREFERRED STOCK |1,500| 750|
----------------------------------------
|Total |2,000|2,000|
----------------------------------------
You now have 2 LEGIT possibilities, in my opinion.
Answer 843012 by greenestamps(12490)  on 2024-02-01 21:21:46 (Show Source):
You can put this solution on YOUR website!
Required information is missing from your post.
With only the given information, it is impossible to find the answer. There are many possible combinations of bonds, preferred stock, and common stock that give a total of 2000 shares at a total cost of $200,000.
Here is what we can do with the problem as stated....
x = # bonds
y = # shares of preferred stock
z = # shares of common stock
The given conditions give us these two equations:
(1)  the total number of shares was 2000
(2)  the total cost was $200,000
With three variables and only two equations, we can only find a family of solutions. The number of solutions is limited by the fact that x, y, and z must be whole numbers.
Simplify (2) (divide everything by 25):
(3) 
Use (1) and (3) to eliminate z; then solve the resulting equation for one of the variables:
(4) 
That gives us an expression for y in terms of x. Now use (1) and (4) to find an expression for z in terms of x:
Now we have expressions for two of the variables in terms of the third, so we can define parametric equations to find different solutions to the problem.
The expression for y in terms of x tells us that any whole number value of x will yield an integer number value for y; and since y must be a whole number (not a negative integer), the expression also tells us the maximum value for x is 400.
The expression for z will not yield a whole number value for any whole number value for x. To find all the whole number solutions to the problem, we could do some further mathematics; I won't go that path, since we already know we can't solve the problem as posted. Instead we can find a few possible answers to the problem by trying different values for x and finding ones that yield whole number values for both y and z.
x y=15(400-x) z=14x-4000
------------------------------------
400 15(0)=0 5600-4000=1600
300 15(100)=1500 4200-4000=200
350 15(50)=750 4900-4000=900
There are 3 combinations that satisfy the given conditions; there are others.
To find a single solution, we would have to have another piece of information. For example, seeing the second possible solution in the list above, we could find that single solution if the given information said either (a) the number of shares of preferred stock was 5 times the number of bonds, or (b) the number of bonds was 1.5 times the number of shares of common stock.
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The student replied in a "thank you" note to me that the answer was 200 shares of common stock, with 1500 shares of preferred stock and 300 bonds. He also stated that he didn't know where they cam up with 200 shares of common stock.
As I showed in my response, that is one of many possible solutions; it is in fact one of the specific solutions I showed.
The fact remains that, with the problem as presented in the student's post, a single solution to the problem is not possible.
Perhaps the student will see this note and look at the given information again, finding that a required piece of information was not provided in his post.
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Trigonometry-basics/1206023: Given: (3i-7)
A. Write in polar form showing work.
B. What is the Cartesian form ?
2 solutions
Answer 843208 by math_tutor2020(3057) on 2024-02-10 12:36:14 (Show Source):
You can put this solution on YOUR website!
I'll discuss a different way to find the angle theta.
tan(theta) = y/x
theta = arctan(y/x)
theta = arctan(3/(-7))
theta = -23.19859 degrees approximately
An angle coterminal to this is roughly -23.19859+360 = 336.80141 degrees
This places the angle in quadrant Q4 (southeast), but it should be in quadrant Q2 (northwest) as the other tutor illustrates.
We must subtract off 180 to get the angle pointed in the right direction.
336.80141 - 180 = 156.80141
Answer 843199 by Edwin McCravy(19626) on 2024-02-09 21:47:03 (Show Source):
You can put this solution on YOUR website! Given: (3i-7)
A. Write in polar form showing work.
First write the cartesian form -7+3i, which is simply (-7,3).
Then realize that it means the vector whose tail is at the origin (0,0)
and whose tip (pointed end) at the point (-7,3)
[In fact that same vector can placed with its tail at any point (a,b),
and its tip, (pointed end) at (a-7,b+3).]
For convenience we place it with its tail at the origin.
We find its magnitude (its length), r, by drawing a line perpendicular
to the x-axis:
and using the Pythagorean theorem.
Next we find the argument or the angle θ, which swings around
counter-clockwise from the right side of the x-axis, indicated by the
blue arc.
We calculate θ by first finding the tangent of its reference angle,
     , rounding off.
But we know that θ is in QII, we subtract from 180o.
The polar form is 
Your teacher might prefer the angle to be in radians instead of degrees,
if so, convert to radians.
Edwin
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Probability-and-statistics/1206029: A plumber tells you she will arrive between 8 am and noon. Assume any time in this interval is an equally likely arrival time. Let X= arrival time (minutes after 8 am).
State the distribution that X follows in proper mathematical notation.
Find the probability density function f(x) and sketch a graph of it in your notes.
What is the probability she arrives before 9 am? Pose this question with mathematical notation and compute the answer.
What is the probability she arrives between 10:15 and 11 am? Pose this question with mathematical notation and compute the answer.
It is now 9:30. What is the probability that she arrives in the next 30 minutes? Pose this question with mathematical notation and compute the answer.
1 solutions
Answer 843205 by ikleyn(49985)  on 2024-02-10 10:51:49 (Show Source):
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Matrices-and-determiminant/1167786: A soap manufacturer decides to spend 60000 rupees on radio, magazine and tv advertising. if he spends as much on tv advertising as on magazines and radio together, and the amount spen on magazines and tv combined equals five times that spent on radio, what is the amount to spent on each type of advertising.using gauss elimination method?
4 solutions
Answer 843201 by greenestamps(12490) on 2024-02-09 22:33:33 (Show Source):
You can put this solution on YOUR website!
It is usually pointless to ask to see a solution using Gauss-Jordan elimination, since there are always an endless number of different paths to the solution using that process.
However, the numbers in this problem are so nice that it can be helpful to solve the problem that way, in order to point out some of the things you can do in general using that process.
So let's do it....
m = amount spent on advertising in magazines
r = amount spent on advertising on radio
t = amount spent of advertising on TV
The given information leads directly to these three equations:
(1) m+r+t = 60000
(2) m+r = t
(3) m+t = 5r
Put these three equations in the required form:
m+r+t = 60000
m+r-t = 0
m-5r+t = 0
And build the initial matrix using those equations:
We first need a "1" in row 1, column 1; we already have that.
Next we need "0"s in the rest of column 1. Since the first column is all "1"s, that is easily accomplished. Replace row 2 with (row 2 minus row 1); replace row 3 with (row 3 minus row 1):
Simplify rows 2 and 3 by dividing by the greatest common factor in each row:
We want "1"s on the main diagonal and "0"s everywhere below the main diagonal; we can do that simply by switching rows 2 and 3:
To finish, we want "0"s above the main diagonal; with this simple example that is easy: replace row 1 with (row 1 minus row 2 minus row 3):
We have the solution.
ANSWER:
m (magazines) 20,000 rupees
r (radio) 10,000 rupees
t (television) 30,000 rupees
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NOTE!...
While this problem lends itself to an easy solution using Gauss-Jordan elimination, it is also very easily solved by standard algebraic methods.
However, the student will get by far the most benefit by working the problem using logical reasoning and simple arithmetic, as demonstrated in the response from tutor @ikleyn.
Answer 843192 by josgarithmetic(39158) on 2024-02-09 13:13:01 (Show Source):
You can put this solution on YOUR website!
RADIO r
MAGAZINE m
TV r+m
TOTAL 60000
--
"gaussian elimination method?"
2 2 60000
-4 2 0
2 2 60000
4 -2 0
1 1 30000
2 -1 0
ADD these.....
3 0 30000
1 0 10000
This means r=10000
Answer 843190 by ikleyn(49985) on 2024-02-09 11:27:58 (Show Source):
You can put this solution on YOUR website! .
A soap manufacturer decides to spend 60000 rupees on radio, magazine and tv advertising.
if he spends as much on tv advertising as on magazines and radio together,
and the amount spent on magazines and tv combined equals five times that spent on radio,
what is the amount to spent on each type of advertising. using gauss elimination method?
~~~~~~~~~~~~~~~~~~~~~
It is easy to solve this problem MENTALLY.
I will show you how to do it. Watch attentively every my step.
(1) Since the soap manufacturer spent as much on tv advertising as on magazines and radio together,
we may conclude that he spent exactly half of the 60000 rupees on tv advertising.
Thus, he spent exactly 30000 rupees on tv advertising.
(2) Next, since the amount spent on magazines and tv combined equals five times that spent on radio,
we may conclude that he spent 1/6 of the 60000 rupees, or 10000 rupees, on radio advertising.
(3) The rest, 60000 - 30000 - 10000 = 20000 rupees was spent for magazine advertising.
Solved.
Answer 843186 by mananth(16751)  on 2024-02-09 05:56:06 (Show Source):
You can put this solution on YOUR website!
A soap manufacturer decides to spend 60000 rupees on radio, magazine and tv advertising. if he spends as much on tv advertising as on magazines and radio together, and the amount spen on magazines and tv combined equals five times that spent on radio, what is the amount to spent on each type of advertising.using gauss elimination method?
A soap manufacturer decides to spend 60000 rupees on radio, magazine and tv advertising.
Let expense on TV be t
Let expense on magazines be m
Let expense on radio be r
t= m+r......................1
m+t = 5r...................2
m+t+r=60000...........3
substitute t in 2
m+t=5r
m+m+r=5r
2m =4r
m=2r
Now m+t=5r
but m=2r
substitute m
2r+t=5r
t=3r
m+t+r=60000...........3
2r+3r+r=60000
6r=60000
r=10000
t=3r
t= 3*10000=30000
m=2r
m=2*10000=20000
you conclude
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Miscellaneous_Word_Problems/1206026: A square corner lot was purchased, and owner intended to construct a house. Owner was to lay a sidewalk facing both streets. 61 blocks of stone were used, each a yard square. Another person offered to purchase the lot at a profit of $100 a sq. ft. Owner accepted offer. Determine amt. he made.
9 sq ft. = 1 sq. yd.
Unsure how to solve.
2 solutions
Answer 843200 by greenestamps(12490) on 2024-02-09 22:07:46 (Show Source):
You can put this solution on YOUR website!
The corner lot is square, and 61 blocks each a yard square were used to lay a sidewalk facing both streets. That means 30 of those blocks were facing each street, with the 61st block on the corner.
So, assuming the sidewalk was not part of the lot, the lot was 30 yards square, or 90 feet square, making the area of the lot 90*90 = 8100 square feet.
His profit on selling the lot was $100 per square foot, which is 8100*$100 = $810000.
ANSWER: $810,000
Note if the lot included the area of the sidewalk, simply do the same calculations using 31 yards = 93 feet instead of 90 feet for the side length of the lot....
Answer 843194 by josgarithmetic(39158) on 2024-02-09 15:40:01 (Show Source):
You can put this solution on YOUR website! This problem is missing the price for a block of stone.
Just for example, block of stone price is some x dollars per square yard.
Offer is some y total payment.
OWNER PAY
ACCEPTED OFFER
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