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Car A and car B are traveling in the same direction. Car A is traveling at a constant velocity of 60 mph and car B is traveling at a constant velocity of 75mph. If car A reaches mile marker M 20 minutes before car B, how much time will elapse between the time that car B reaches mile marker M and the time that car B reaches car A?
:
During the 20 min that elapses for car B to get to the marker, car A will have traveled:
* 60 = 20 miles (from the marker)
:
Let t = time for the car B to catch car A
:
Write a distance equation Distance = speed * time
:
Car B distance = Car A distance + 20 miles
75t = 60t + 20
75t - 60t = 20
15t = 20
t =
t = = 1 hrs
:
Another easy way:
Using relative speed. 75 - 60 = 15 mph
Assume that Car A is stationary and car B is traveling at 15 mph
:
How long will it take Car B to travel 20 mi at 15 mph?
time = = 1 hrs

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a road side stand bought 11 large baskets of eggs from a farmer and sold 39 small baskets of eggs, which hold fewer than a dozen. there were 19 eggs left over. how many eggs does a large basket hold?
;
Let y = no. of eggs in a large basket
Let x = no. of eggs in a small basket
:
The equation:
11y = 39x + 19
y =
We know that y has to be an integer and (39x+19) has to be a multiple of 11
Not many values of x less than 12, will fit the requirement
:
A few minutes with a calc and you come up with x = 6
y =
y = 23 eggs in a large basket (6 in a small basket)

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a contractor has a 24-pound mixture that is no-fourth cement, and three-fourths sand. How much of a mixture that is half cement and half sand needs to be added to produce a mixture that is one-third cement?
:
We can write an equation based on the amt of cement
;
Let x = amt of half cement mixture required
:
(24) + x = (x+24)
Multiply equation by 12 to get rid of the denominators
:
3(24) + 6x = 4(x+24)
:
72 + 6x = 4x + 96
:
6x - 4x = 96 - 72
:
2x = 24
x =
x = 12 lb of the half cement mixture
;
:
we can check solution base on the amt of sand:
(24) + (12) = (12+24)
18 + 6 = *36

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Write a variable expression for "7 divided by the sum of x and 5"
;
Wouldn't it just be:
?

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A and B working together can do a job in 24 hours. After A worked alone for 7 hours, B joined him and together they finished the rest of the job in 20 hours. How long would it take each working alone to do the job?
;
Let a = time required when A works alone
Let b = time required when B works alone
:
Let the completed job = 1
:
write an equation for each scenario :
"A and B working together can do a job in 24 hours."
+ = 1
Multiply equation by ab to get rid of the denominators, results
24b + 24a = ab
24a = ab - 24b
24a = b(a-24)
= b
:
"After A worked alone for 7 hours, B joined him and together they finished the rest of the job in 20 hours.
That means a worked 27 hrs
+ = 1
Multiply equation by ab to get rid of the denominators, results
27b + 20a = ab
20a = ab - 27b
20a = b(a-27)
= b
:
Therefore:
=
Cross multiply:
24a(a-27) = 20a(a-24)
:
24a^2 - 648a = 20a^2 - 480a
:
24a^2 - 20a^2 -648a + 480a = 0
:
4a^2 - 168a = 0
:
4a(a - 42) = 0
:
a = +42 A's hrs alone
;
Use + = 1 to find b, substitute 42 for a
+ = 1
Multiply equation by 42b
24b + 42(24) = 42b
:
1008 = 42b - 24b
:
1008 = 18b
b =
b = 56 hrs is B's time alone
;
:
You can check solution using a calc on both equations, a=42, b=56

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the four oldest people in Golden City have lived 384 years. the difference in ages for the youngest and the next oldest is 14. the next youngest is 3 years older than the youngest. the oldest is 20 years older than the average of the next oldest and youngest. How old is each?
:
Let the 4 people be w,x,y,& z (youngest to oldest)
:
Write an equation for each statement
:
"the four oldest people in Golden City have lived 384 years."
w + x + y + z = 384
:
"the difference in ages for the youngest and the next oldest is 14."
y - w = 14
y = w + 14
:
"the next youngest is 3 years older than the youngest.
x = w + 3
:
the oldest is 20 years older than the average of the next oldest and youngest.
z = + 20
multiply equation by 2 to get rid of the denominator
2z = w + y + 2(20)
2z = w + y + 40
Replace y with (w+14)
2z = w + (w+14) + 40
2z = 2w + 54
Divide equation by 2:
z = w + 27
:
How old is each?
:
Replace x, y, and z in the total equation with the above equations:
w + (w+3) + (w+14) + (w+27) = 384
:
4w + 44 = 384
:
4w = 384 - 44
:
4w = 340
w =
w = 85 yrs is the youngest oldie
then
x = 85 +3
x = 88 yrs is the next oldie
and
y = 85 + 14
y = 99 yrs
and
z = 85 + 27
z = 112 yrs is real old guy
;
:
Check solution by finding the total
85 + 88 + 99 + 112 = 84

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A rectangular garden has dimensions of 18 feet by 13 feet. A gravel path of uniform width is to be built around the garden. How wide can the path be if there is enough gravel for 516 square feet?
:
Draw a rectangle representing the 18 by 13 ft garden, then draw a larger rectangle
around that one, enclosing the path around the garden.
:
Label the width of the path as x
It will be apparent that the overall dimensions will be (18+2x) by (13+2x)
FOIL this to get the overall area:
(18+2x)*(13+2x) = 234 + 36x + 26x + 4x^2 = 234 + 62x + 4x^2
:
Garden area; 18 * 13 = 234
:
The equation:
Overall area - garden area = path area (given as 516 sq/ft)
(4x^2 + 62x + 234) - 234 = 516
:
4x^2 + 62x - 516 = 0; our old friend, the quadratic equation
Simply divide equation by 2:
2x^2 + 31x - 258 = 0
Factor this to:
(2x + 43)(x - 6) = 0
Positive solution
x = +6 ft is the width of the path
:
:
We can check this: overall dimensions will be (18+12) by (13+12)
(30*25) - 234 =
750 - 234 = 516, confirms our solution
:
:
Did this make sense? Any questions?

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Two ranchers sold a herd of cattle and received as many dollars for each animal as there are animals in the herd. If there were two animals, they received $2 per animal; if there were three animals, they received $3 per animal; etc. As an example, if there were eight animals in the herd, the value of the herd would be $64 ($8 per animal times 8 animals in the herd). We’re told that they started with no more than 20 cattle.
With the money, they bought as many sheep as they could at $10 per head, and a goat with the remainder (less than $10). Finally they divided the animals between them. There was however, an odd number of sheep. So, one rancher took the extra sheep and the other the goat. The rancher who got the goat was given his friends new pocket knife as compensation. If this was to be an equal monetary division, what was the value of the pocketknife.
:
We know a couple of things,
the number of sheep received from the sale has to be odd.
the number of sheep $ value, has to leave a remainder, to buy the goat with
:
Make a table using the value of the cattle = x, (has to be 20 or less)
Let y = number of sheep obtained from this value
Make a table from the equation y = x^2/10
x | y
--------
4 |1.6*
5 |2.5
6 |3.6*
7 |4.9
8 |6.4
9 |8.1
10 |10
11 |12.1
12 |14.4
13 |16.9
14 |19.6*
15 |22.5
16 |25.6*
:
*It's interesting to note that when y has has a odd whole number,
the decimal is always .6
Since we have to have an odd number of sheep as given by the problem, we can
assume the goat cost $6, therefore the knife value is $4
:
the goat and the knife count as 1 ea $10 sheep, making each man get the same monetary value
;
Did this make sense??

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find the total amount due for each simple interest loan for $6100 at 5.7% interest for 4 years 9 months.
:
Find the interest due for 4.75 yrs
:
i = p*r*t
p = 6100
r = .057
t = 4.75
:
i = 6100 * .057 * 4.75
i = 1651.58
add to the principal
1651.58 + 6100 = $7,751.58

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A man in a canoe is paddling upstream. He passes a floating log (going downstream) at the Elk's Lodge. He continues to paddle for 90 more minutes, before turning around to go downstream. The canoer reaches the log again at the sewer plant which is exactly 1 mile from the Elk's lodge. At what rate is the river flowing?
:
Here's what I think:
:
The trip away from log will take the same time as the trip back to the log
The speed is the same both ways in relation to the water.
:
2 * 90 min = 3 hr
During this time the log travels 1 mi in relation to the land.
Current = mph

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In three years, Alex will be 3 times as old as Precy.
A year ago, Alex was 7 times as old as Precy. How old are they now?
:
Let A and P = their present ages
:
Write an equation for each statement, simplify:
:
"In three years, Alex will be 3 times as old as Precy."
A+3 = 3(P+3)
A + 3 = 3P + 9
A = 3P + 9 - 3
A = 3P + 6
:
"A year ago, Alex was 7 times as old as Precy."
A-1 = 7(P-1)
A - 1 = 7P - 7
A = 7P - 7 + 1
A = 7p - 6
:
How old are they now?
:
Substitute (3P+6) for A in the above equation
3P + 6 = 7p - 6
6 + 6 = 7P - 3P
12 = 4P
P =
P = 3 yrs old Precy's present age
and
A = 3(3)+ 6
A = 15 yrs old is Alex's present age
;
:
Check the solution in the statement:
"A year ago, Alex was 7 times as old as Precy."
15 - 1 = 7(3-1)

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Create a real-life situation that fits into the equation (x+4)(x-7) = 0 and express the situation as the same equation.
:
A simple problems would be. You want a rectangular area of 28 sq/ft
You want the length to be 3 ft longer than the width, Find the dimensions:
;
Let x = length
then
(x-3) = width
and
Area = x(x-3) = x^2 - 3x
:
Area equation:
x^2 - 3x = 28
:
A quadratic equation
x^2 -3x - 28 = 0
:
factor this to:
(x-7)(x+4) = 0
:
Positive solution:
x = +7 is the length

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We use the discriminant:
with the equation in the form ax^2 + bx + c;
Use the values for a, b, c in:
b^2 - 4*a*c =
The rules for the discriminant are are:
b^2 - 4*a*c = 0; one real number solution, a double root
b^2 - 4*a*c > 0; two real number solutions, that are not equal
b^2 - 4*a*c < 0; two complex number solutions

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the length of a rectangular swimming pool is twice its width. The pool is surrounded by a cement walk 4 feet wide. If the area of the walk is 748 ft^2, determine the dimensions of the pool.
:
Let x = the width of the pool
then
2x = the length
and
2x^2 = area of the pool
:
Overall area including the walk = (x+8)*(2x+8) = 2x^2 + 24x + 64
:
The equation:
Overall area - pool area = 748 sq/ft (walkway area)
(2x^2 + 24x + 64) - 2x^2 = 748
:
24x = 748 - 64
:
24x = 684
x =
x = 28.5 ft width of the pool
and
2(28.5) = 57 ft, length of the pool
:
:
Check solution:
Overall area = 65 * 36.5 = 2372.5
---Pool area = 57 * 28.5 = 1624.5
----------------------------------
leaving the walkway area = 748

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A cyclist rode 40 mi. before having a flat tire and then walking 5 mi to a service station. The cycling rate was four times the walking rate. The time spent cycling and walking was 5 h. Find the rate at which the cyclist was riding.
:
Let x = the walking rate
then
4x = the riding rate
:
Write a time equation: Time =
:
Riding time + walking time = 5 hrs
+ = 5
Reduce the fraction:
+ = 5
= 5
Multiply both sides by x
5x = 15
x =
x = 3 mph is the walking rate
then
4(3) = 12 mph is the riding rate
;
:
Check solution
+ =
3 + 1 = 5 hrs

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A plane which can fly 100 miles an hour in still air can fly 500 miles with a wind which is blowing at a certain rate in 5/8 of the time it would require to fly 500 miles against a wind blowing at the same rate. What was the rate of the wind?
:
Let x = the rate of the wind
then
(100-x) = speed against the wind
and
(100+x) = speed with the wind
:
Write a time equation: Time =
:
With the wind time = Against time *
= *
=
=
Cross multiply:
2500(100+x) = 500(800-8x)
:
250000 + 2500x = 400000 - 4000x
;
2500x + 4000x = 400000 - 250000
:
6500x = 150000
x =
x = 23.1 mph, the rate of the wind
:
:
Check solution, find the time for each trip
= 4.06 hrs
* = 4.06 hr

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Please Simplify:
3x^2-22x+7
:
You can factor this to:
(3x - 1)(x - 7)

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Treat it like a triangular prism, the formula for that is:
V = * L * W * H
:
Your attic:
V = * 7 * 6.5 * 2.8
:
V = 63.7 cu meters

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Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle.
:
Let s = his speed
:
Write a time equation: Time =
:
= + 1
Multiply equation by s(s+10), results:
:
200(s+10) = 200s + s(s+10)
:
200s + 2000 = 200s + s^2 + 10s
:
Subtract 200s from both sides. Arrange as a quadratic equation:
s^2 + 10s - 2000 = 0
Factor:
(s+50)(s-40) = 0
The positive solution
s = 40 mph
:
:
Check solution
200/40 = 200/50 + 1

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Car A and car B are traveling in the same direction. Car A is traveling at a constant velocity of 60 mph. and car B is traveling at a constant velocity of 75mph.
:
If car A reaches mile marker M 4 minutes after car B, how much time will elapse between the time that car A reaches mile marker M, and the time that the cars are 20 miles apart?
:
When car A reaches the M marker, car B will have traveled
* 75 = 5 miles
The relative speed between the two cars: 75 - 60 = 15 mph
:
Think of car A being stationary, car B is 5 mi away traveling at 15 mph
:
How long would it take Car B to travel another 15 mi at 15 mph?
:
They will be 20 mi apart in 1 hr from the time Car A is at the marker
:

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The length of a rectangle must be 4 feet more than the width, and the perimeter must be at least 32 feet. What is the range of values for the width?
:
Let x = width
then
(x+4) = length
:
Perimeter
2(x+4) + 2x => 32
;
2x + 8 + 2x => 32
:
4x => 32 - 8
:
x =>
x => 6 ft width

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Find the total grazing area of a goat G, The animal is tied to a corner of a 40'x40' barn, by an 80' rope. One of the sides of the barn is extended by a fence. assume that the area is grass everywhere except inside of the barn.
;
The side that is extended by the fence will make a difference
:
If the fence is opposite the corner, where the rope is tied, then is will not effect it.
:
The larger area is of circle with 80' radius
A =
A = 15079.6 sq/ft
and
2 each circles with a radius of 40'
A =
A = 2513.3 sq/ft
:
Total: 15079.6 + 2513.3 = 17,593 sq/ft

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All these values would put 0 as the denominator, a big no-no in algebra

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i) Problem: A sheet metal worker is planning to make an open-top box by cutting equal squares (x-in. by x-in.) from the corners of a 10-in. by 14-in. piece of copper. A second box is to be made in the same manner from an 8-in. by 10-in. piece of aluminum, but its height is to be one-half that of the first box.
:
(1)Find a polynomial function for the volume of each box.
:
1st box dimensions: (10-2x) by (14-2x) by x
Vol = x*(10-2x)*(14-2x)
FOIL
f(x) = x(140 - 48x + 4x^2)
f(x) = 4x^3 - 48x^2 + 140x
:
2nd box cut out will be .5x in. by .5x in.,(height 1/2 the 1st box), therefore:
the dimensions will be:
(10-x) by (8-x) by .5x
Vol = .5x*(10-x)*(8-x)
FOIL
f(x) = .5x(80 - 18x + x^2)
f(x) = .5x^3 - 9x^2 + 40x
:
:
(2)Find the values of x for which the copper box is 72 cubic in. larger than the aluminum box.
:
Big box vol - small box vol = 72 cu in
(4x^3 - 48x^2 + 140x) - (.5x^3 - 9x^2 + 40x) = 72
Remove brackets
4x^3 - 48x^2 + 140x - .5x^3 + 9x^2 - 40x = 72
Group like terms
4x^3 - .5x^3 - 48x^2 + 9x^2 + 140x - 40x - 72 = 0
:
3.5x^3 - 39x^2 + 100x - 72 = 0
:
Solve this by graphing y = 3.5x^3 - 39x^2 + 100x - 72

:
The integer solution x = 2; & x ~ 1.3; the third solution x ~ 7.8 isn't reasonable
:
(3)Write the difference between the two volumes (d) as a function of x.
d(x) = (4x^3 - 48x^2 + 140x) - (.5x^3 - 9x^2 + 40x)
which is
d(x) = 3.5x^3 - 39x^2 + 100x
:
(4)Find d for x=1.5
d(x) = 3.5(1.5^3) - 39(1.5^2) + 100(1.5)
d(x) = 3.5(3.375) - 39(2.25) + 150
d(x) = 11.8125 - 87.75 + 150
d(x) = 74.0625
:
(5)For what value of x is the difference between the two volumes the largest?
Graph d(x) = 3.5x^3 - 39x^2 + 100x:

:
Greatest difference appears to be at x = 1.7 which is a difference of about 74.5

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1.) One number is 3 more than the other. Their sum is 7. Find the numbers.
Let x = "one number
then
(x-3) = "other number
:
"Their sum is 7"
x + (x-3) = 7
2x = 7 + 3
x =
x = 5
:
Check: 5 - 3 = 2: 5 + 2 = 7
:
:
2.) The denominator of a fraction is twice the numerator. If the sum of the
terms is 3, what is the fraction?
Let x = the numerator
then
2x = denominator
Obviously, if x + 2x = 3, then x = 1 and the fraction is
:
:
3.) The sum of two numbers is 165 and their difference is 29. Find the
numbers.
x + y = 165
x - y = 29
------------adding eliminates y:
2x + 0 = 194
x =
x = 97
I'll let you find y
:
:
4.) If the numerator of a certain fraction is increased by 6 and its
denominator is decreased by 5, the resulting fraction is equal to 3/4.
=
:
If the reciprocal of the original fraction is decreased by 1, the
resulting fraction is 16/9.
- 1 =
= + 1
= +
=
Find the original fraction.
:

;
Check solution in the 1st statement:
=
= = ; confirms our solutions
:

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Walt made an extra 9000 last year from a part time job. He invested part of the money at 9% and the rest at 8%. He made a total of $770.00 in interest. How much was invested at 8%?
:
Let x = amt invested at 8%
The total was given as $9000, therefore:
(9000-x) = amt invested at 9%
:
Write an interest equation:
:
interest at 8% + interest at 9% = $770
.08x + .09(9000-x) = 770
Multiply what's inside the brackets
.08x + 810 - .09x = 770
:
.08x - .09x = 770 - 810
:
-.01x = -40
x =
x = $4000 invested at 8%
:
:
Check solution ($5000 invested at 9%)
.08(4000) + .09(5000) =
320 + 450 = 770; confirms our solution

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1.
Cube each term
:You multiply when you raise one power to another power

:
2.

; a negative raised to an even power will be positive
:
3.


:
4.


:
5.

; a negative raised to an odd power will be negative
: add exponents of like terms when you multiply

:
This is the method, check each step to ensure I did not make a dumb math mistake here

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Brenda made 46 and 59 on her first two Biology tests and has one test remaining. The average on the three tests must be at least 65 for Brenda to pass the course. Write an inequality to represent this situation using s for the score on the last test.
=> 65
Multiply equation by 3 to get rid of the denominator
46 + 59 + s => 3(65)
:
105 + s => 195
:
s => 195 - 105
:
s => 90 score required to pass
:
:
Check solution
= 65

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an express and local train leave Gray's Lake at 3 P.M. and head for Chicago 50 miles away. The express travels twice as fast as the local, and arrives 1 hour ahead of it. Find the speed of each train.
:
The time they left is irrelevant to the problem
:
Let s = speed of the local
then
2s = speed of the express
:
Write a time equation: Time =
:
Express time + 1 hr = Local time
+ 1 =
Get rid of the denominators, mult equation by 2s; results:
50 + 2s = 2(50)
2s = 100 - 50
2s = 50
s =
s = 25 mph speed of the local
then
50 mph speed of the express
;
;
Check solution
+ 1 =
1 + 1 = 2

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Ely stands next to his cousin Teri. He shoots his bb-gun at a straight horizontal angle. Teri records the time it takes from the moment Ely fires the bb-gun until the sounds of the ping(the bb hitting the can) to return, .25 seconds. The distance is 20 yards. The speed of sound is 1140 ft per sec. How fast does the bb travel? How far did the bb fall? The equation he gave us h=-16t^2+vt+h
:
Let x = velocity of the bb
:
Write a time equation: Time = dist/speed
:
Distance for the sound and the bb = 20 yds which = 60 ft (since we're dealing in ft/sec}
:
Return sound time + bb transit time = .25 sec
+ = .25
:
Multiply equation 1140x and you have:
60x + 1140(60) = 1140x * .25
:
60x + 68400 = 285x
:
68400 = 285x - 60x
:
68400 = 225x
x =
x = 304 ft/sec, velocity of the bb gun
:
:
How far did the bb fall:
Find the transit time of the bb
= .197368 sec to travel 60 ft
The horizontal velocity does not affect the fall distance, only the transit time and gravity (-16t^2)
-16*.197368^2 = -.03895 ft or about .467 inches

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Write an equation for each statement:
:
"Bob is half as old as Tom."
B = T
or
T = 2B
:
"One year ago Bob age was two fifths of what Tom's age will be a year from now."
B - 1 = (T + 1)
Multiply equation by 5 to get rid of the denominator, results:
5(B - 1) = 2(T + 1)
:
5B - 5 = 2T + 2
:
5B = 2T + 2 + 5
:
5B = 2T + 7
:
Can you figure out Bob's age and Tom's age.
:
Substitute 2B for T in the above equation
5B = 2(2B) + 7
5B = 4B + 7
5B - 4B = 7
B = 7 yrs is Bob's age
Then
T = 2(7)
T = 14 yrs is Tom's age
:
:
Check solution using the statement:
"One year ago Bob age was two fifths of what Tom's age will be a year from now."
7 - 1 = (14 + 1)
6 = *15