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two angles are complementary. one angle is 12 degree larger than the other. using two variables x and y, find the size of each angle by solving a system of equations.
.
Two angles are complementary if the sum of the two angles equals 90 deg.
.
Let x = one angle
and y = second angle
.
equation 1:
x + y = 90
equation 2:
y = x+12
.
Substitute the value of y from equation 2 into equation 1 and solve for x:
x + y = 90
x + x+12 = 90
2x + 12 = 90
2x = 78
x = 39 degrees
.
Substitute the above into equation 2 and solve for y:
y = x+12
y = 39+12
y = 51 degrees
.
Angles are 39 and 51 degrees.

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If a parade 2mi long is proceeding at 3 mph, how long will it take a runner, jogging at 6 mph, to travel from the front of the parade to the end of the parade?
.
Apply the distance formula:
d = rt
where
d is distance
r is rate or speed
t is time
.
Let x = time it takes jogger to get to end of parade
then
3x + 6x = 2
9x = 2
x = 2/9 = 1/3 hours
or
x = 20 minutes

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Find an equation of the the line satisfying the given conditions.
Through (2, 8); perpendicular to 9x + 7y = 74
.
First, find the slope of:
9x + 7y = 74
Do this by putting it into the "slope-intercept" form:
7y = -9x + 74
y = (-9/7)x + 74/7
Therefore, slope = -9/7
.
If a line is perpendicular, we need the negative reciprocal of the slope above:
Our new slope is then 7/9
Use the slope (7/9) and the given point (2,8) and plug it into the "point-slope" form:
y-y1 = m(x-x1)
y-8 = 7/9(x-2)
y-8 = 7/9(x) - 14/9
y = 7/9(x) - 14/9 + 8
y = 7/9(x) - 14/9 + 72/9
y = 7/9(x) + 58/9 (slope-intercept form of new line)

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Determine whether the given ordered pair is a solution of the given equation.
(x - 4)^2 + (y + 7)^2 = 45; (1, -1)
.
Plug in the given ordered pair: (x,y) = (1,-1) into
(x - 4)^2 + (y + 7)^2 = 45
(1 - 4)^2 + (-1 + 7)^2 = 45
(-3)^2 + (6)^2 = 45
9 + 36 = 45
45 = 45 (checks)
.
So, yes, (1,-1) is a solution

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Write an equation in standard form for a line passing through the pair of points.
(2, -2) and (0, 3)
.
First, find the slope between the two points:
m (slope) = (y2-y1)/(x2-x1)
m (slope) = (3-(-2))/(0-2)
m (slope) = (3+2)/(0-2)
m (slope) = (5)/(-2)
m (slope) = -5/2
.
Using the slope above and one point (0,3) plug it into the point slope form:
y-y1 = m(x-x1)
y-3 = -5/2(x-0)
y-3 = -5/2(x)
5/2(x) + y - 3 = 0 (this is what they're looking for)

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Find the slope of the line, if it is defined.
Through (-3, -9) and (5, 4)
.
m (slope) = (y2-y1)/(x2-x1)
m (slope) = (4-(-9))/(5-(-3))
m (slope) = (4+9)/(5+3)
m (slope) = (13)/(8)
m (slope) = 13/8

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Find the slope of the line, if it is defined.
Through (7, -4) and (3, 8)
m (slope) = (y2-y1)/(x2-x1)
m (slope) = (8-(-4))/(3-7)
m (slope) = (8+4)/(3-7)
m (slope) = (12)/(-3)
m (slope) = -4

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Maria has $169 in ones,fives, and tens. She has twice as many one-dollar bills as she has five dollar bills, and five more ten-dollar bills than five-dollar bills. How many of each type bill does she have?
.
Let x = number of five dollar bills
then
2x = number of one dollar bills
x+5 = number of ten dollar bills
.
2x + 5x + 10(x+5) = 169
2x + 5x + 10x+50 = 169
17x+50 = 169
17x = 119
x = 7 (number of five dollar bills)
.
One dollar bills:
2x = 2(7) = 14
.
Ten dollar bills:
x+5 = 7+5 = 12

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In 1920, the record for a certain race was 45.4 sec. In 1990, it was 44.0 sec.
Let R(t)= the record in the race and t= the number of the year since 1920.
.
a) Find a linear function that the date
Think of the x-axis representing the year and the y-axis representing the "record-time" in seconds.
The problem, then, really gives you two data points: (remember, since 1920)
1920 we get: (0,45.4)
1990 we get: (70,44)
Slope then is
(y2-y1)/(x2-x1) = (44-45.4)/(70-0) = -1.4/70 = -0.02
.
Using the slope and one (70,44) of the two points substitute it into the "point-slope" form:
y-y1 = m(x-x1)
y-44 = -0.02(x-70)
y = -0.02(x-70) + 44
y = -0.02x + 1.4+44
y = -0.02x + 45.4
Replacing y with R(t) and x with t we get:
R(t) = -0.02t + 45.4
.
b) Use the function in (a) to predict the record in 2003 and in 2006.
For 2003:
t = 2003-1920 = 83
R(t) = -0.02t + 45.4
R(t) = -0.02(83) + 45.4
R(t) = -1.66 + 45.4
R(t) = -1.66 + 45.4
R(t) = 43.74 seconds
.
For 2006:
t = 2006-1920 = 86
R(t) = -0.02t + 45.4
R(t) = -0.02(86) + 45.4
R(t) = -1.66 + 45.4
R(t) = -1.72 + 45.4
R(t) = 43.68 seconds
.
c) Find the year when the record will be 43.56 sec
R(t) = -0.02t + 45.4
43.56 = -0.02t + 45.4
43.56-45.4 = -0.02t
-1.84 = -0.02t
92 = t
Year then = 1920+92 = 2012

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A rectangle is 4 times as long as it is wide. A second rectangle is 5 centimeters longer and 2 centimeters wider than the first. The area of the second rectangle is 270 square centimeters greater than the first. What are the demensions of the original rectangle?
.
Let w = width of original rectangle
then
4w = length of original rectangle
.
(w+2)(4w+5) = w(4w) + 270 (if you don't see this, write back)
4w^2+5w+8w+10 = 4w^2 + 270
4w^2+13w+10 = 4w^2 + 270
13w+10 = 270
13w = 260
w = 20 cm (width of original rectangle)
.
Length:
4w = 4(20) = 80 cm (length of original rectangle)

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Employees of a discount appliance store receive an additional 20% off of the lowest price on an item. If an employee purchases a dishwasher during a 15% off sale, how much will he pay if the dishwasher originally cost $450?
.
Original cost: $450
.
15% off original cost:
450 - .15(450)
= 450(1-.15)
= 450(.85)
= $382.5
.
addition 20% discount for employee:
382.5 - .20(382.5)
= 382.5(1-.20)
= 382.5(.80)
= $306
.
Answer: D

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Find an equation of the line satisfying the given conditions.
Through (0,5); m=-(2/3)
.
Plug into the "point-slope" form:
y-y1 = m(x-x1)
.
y-5 = -(2/3)(x-0)
y-5 = -(2/3)x
y = -(2/3)x + 5 (this is what they're looking for)
The above is the "slope-intercept" form.
Where
slope (m) is -(2/3)
and
y-intercept is at (0,5)

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2x-5y=18;(4,2)
.
An "ordered pair" is denoted as (x,y)
So, in this problem, they are asking if
x were equal to 4
and
y were equal to 2
is that a solution for:
2x-5y=18
.
To check,
Plug the given values back into the equation and check:
2x-5y=18
2(4)-5(2)=18
8-10=18
-2 = 18 (NOT true)
.
Therefore (4,2) is NOT a solution.

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No.
Two lines will not intersect if they are parallel.

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The sum of Richard's age and Ruel's age is 60. Nine years ago, Richard was twice as old as Ruel, then how old is Ruel?
.
Let x = Ruel's age
and y = Richard's age
.
From:"The sum of Richard's age and Ruel's age is 60." we get:
x + y = 60 (equation 1)
.
From:"Nine years ago, Richard was twice as old as Ruel" we get:
y-9 = 2(x-9) (equation 2)
.
Solve equation 1 for y:
x + y = 60
y = 60 - x
.
Substitute the above back into equation 2 and solve for x:
y-9 = 2(x-9)
60-x -9 = 2(x-9)
51-x = 2x-18
51 = 3x-18
69 = 3x
23 years = x (Ruel's age)

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.
Multiplying both sides by a common denominator: (x+4)(x-3) we get:





.
x = {-1/2, 4}

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I believe you mean: (correct me if I'm wrong)
.

.
Factoring the denominators we get:

.
Now, find common denominator: (a-9)(a+6)(a-1)
and convert each term to have the same denominator:

.
Combining the two terms:

.


.
Factoring the numerator:

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Investing her bonus. Donna invested her 33,000 bonus and recieved a total of $970 in interest after one year. If part of the money returned 4% and the remainder 2.25% then how much did she invest at each rate?
.
Let x = amount invested at 4%
and y = amount invested at 2.25%
.
From:"Donna invested her 33,000 bonus"
x + y = 33000 (equation 1)
.
From:"a total of $970 in interest after one year"
.04x + .0225y = 970
.
Solving equation 1 for y:
x + y = 33000
y = 33000 -x
.
Substitute the above into equation 2 and solve for x:
.04x + .0225y = 970
.04x + .0225(33000-x) = 970
.04x + 742.5 -.0225x = 970
.04x -.0225x = 227.5
0.0175x = 227.5
x = $13000 (amount invested at 4%)
.
Amount invested at 2.25%
y = 33000 -x = 33000 -13000 = $20000

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can you please help me solve ((( A rectangular solid with length 8 and width3 has bolume 48. Find its height. )))
.
Since the volume on any rectangular solid is length*width*height we have:
.
Let h = height
then
8*3*h = 48
24h = 48
h = 2 (height)

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A rectangular swimming pool is 2 m longer than it is wide. If the width is decreased by 3 m, and the length is increased by 4 m, the area remains the same as the original area, Find the original dimensions of the pool.
.
Let w = width of pool
then
w+2 = length of pool
.
(w-3)(w+2 +4) = w(w+2)
(w-3)(w+6) = w(w+2)
w^2+6w-3w-19 = w^2+2w
w^2+3w-19 = w^2+2w
3w-19 = 2w
w-19 = 0
w = 19 m (width of pool)
.
Length:
w+2 = 19+2 = 21 m (length of pool)
.
Dimensions: 19m x 21m

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The area fo a rectangle is 66ft squared and the diagonal of the rectangle is square root of 157 ft. What are the dimensions of the rectangle.
.
Let x = width
and y = length
.
Since we have two unknowns, we'll need two equations:
Equation 1:
xy = 66
Equation 2:

x^2 + y^2 = 157
.
Solving equation 1 for y:
y = 66/x
Substituting the above into equation 2 we can solve for x:
x^2 + y^2 = 157
x^2 + (66/x)^2 = 157
x^2 + 4356/x^2 = 157
x^4 + 4356 = 157x^2
x^4 - 157x^2 + 4356 = 0
Factoring we get:
(x^2-121)(x^2-36) = 0
x = {11,6}
.
Dimensions of rectangle are 11 feet by 6 feet

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V=1/3BH
Multiplying both sides by 3:
3V=BH
Dividing both sides by B:
3V/B = H

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Volume of any sphere:

So, if r = 3 cm
then
.

.

.

.
cubic cm

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By definition, all sides of a cube are the same.
Therefore, the volume is:
2*2*2 = 8

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a farmer and his son leave a barn at the same time and walk in opposite driections checking a fence line. the son walks at the speed of 3.5km/h and the farmer at 4.0km/h. how much time will have elapsed when the farmer and his son are 2.5km apart
.
Apply the distance formula:
d = rt
where
d is distance
r is rate or speed
t is time
.
Let t = elapsed time
then
3.5t + 4t = 2.5
7.5t = 2.5
t = 2.5/7.5
t = 25/75
t = 1/3 of an hour (or 20 minutes)

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Plot point (-5,8) on the graph
Plot point (3,4) on the graph
Connect the two points with a solid line
The vector starts at A and ends at B, so draw the arrow head at B.

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.
You have to "rationalize" the fraction -- remove all radicals from the denominator.
.

.

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5.8 + [-log(10^-6)-log(1)-2](2.9)= x
5.8 + [-(-6)-0-2](2.9)= x
5.8 + [6-0-2](2.9)= x
5.8 + [4](2.9)= x
5.8 + 11.6= x
5.8 + 11.6= x
17.4=x

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A circle is tangent to the x-axis and to the y-axis. The coordinates of its center are both positive. The area of the circle is 64pie. What are the coordinates of the point of tangency on the y-axis?
.
Area of a circle = (pi)r^2
.
They gave us the area as 64(pi)
.
64(pi) = (pi)r^2
64 = r^2
8 = r (radius)
.
Since the coordinates of the center are both positive, we know it is in the 1st quadrant.
.
point tangent on the y-axis is then (0,8)

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It takes Steve 3 hrs longer to paint a floor than it takes Paul. When working together, it takes them 2 hours. How long would each take to do the job alone?
.
Let p = time it takes for Paul to paint the floor alone
then
p+3 = time it takes for Steve to paint the floor alone
.
2(1/p + 1/(p+3)) = 1
2/p + 2/(p+3) = 1
2(p+3) + 2p = p(p+3)
2p+ 6 + 2p = p^2+3p
8p+ 6 = p^2+3p
6 = p^2-5p
0 = p^2-5p-6
0 = (p-6)(p+1)
p = {6, -1}
We throw out the negative solution leaving us with:
p = 6 hours (time it takes Paul)
.
Steve:
p+3 = 6+3 = 9 hours (time it takes Steve)

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Applying the substitution method:
x-y=5 (equation 1)
3x+y=3 (equation 2)
.
Solve equation 1 for x:
x-y=5
x = y+5
Substitute the above into equation 2 and solve for y:
3x+y=3
3(y+5)+y=3
3y+15+y=3
4y+15=3
4y = -12
y = -3
.
Substitute the above into equation 1 and solve for x:
x-y=5
x-(-3)=5
x+3 = 5
x = 2
.
Solution:
x=2 and y=-3