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12x(Squared) +19x-18= 0
(3x-2)(4x+9) = 0
.
Setting each to zero:
(3x-2)= 0
3x = 2
x = 2/3
and
(4x+9) = 0
4x = -9
x = -9/4
.
x = {-9/4, 2/3}

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A rectangular flower bed 30 yards long by 20 yards wide has a walk of uniform width around it. If the area of the path is 1/4 that of the flower bed, find the width of the path.
.
Let w = width of path
then
Area of flower bed = (30)(20) = 600 sq yards
.
Area of flower bed and walk = (30+2w)(20+2w)
.
Area of path = "area of flower bed and walk" - "area of flower bed"
Area of path = (30+2w)(20+2w) - 600
.
(1/4)600 = (30+2w)(20+2w) - 600
150 = (30+2w)(20+2w) - 600
750 = (30+2w)(20+2w)
750 = 600+60w+40w+4w^2
750 = 4w^2+100w+600
0 = 4w^2+100w-150
0 = 2w^2+50w-75
Using the quadratic equation we get:
w = {1.42, -26.42}
Throw out the negative solution leaves us with:
w = 1.42 yards
.
Details of quadratic to follow:

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Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=3100 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 1.41941090707505, -26.4194109070751. Here's your graph:

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The altitude of a triangle is 3/4 the length of its base. If the altitude were increased by 3 feet and the base decreased by 3 feet, the area would be unchanged. Find the length of the base and altitude.
.
Original triangle:
Let b = length of base
then
(3/4)b = altitude
Area = (1/2)b(3/4)b = (3/8)b^2
.
Changed triangle:
b-3 = length of base
(3/4)b+3 = altitude
Area = (1/2)(b-3)(3/4)b+3 = (3/8)(b-3)b+3
.
Set area equal to each other:
(3/8)b^2 = (3/8)(b-3)b+3
b^2 = (b-3)b+3
b^2 = b^2-3b+3
0 = -3b+3
-3 = -3b
1 feet = b (base of triangle)
.
Altitude:
(3/4)b = (3/4)1 = 3/4 feet (altitude)

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You can solve by one of three methods:
Substitution, addition or matrix.
.
I'll apply the "addition method":
x+ y= 8
51x+8y=193
.
Multiply first equation by -8 to get:
-8x-8y=-64
51x+8y=193
.
Add the two equations together:
-8x-8y=-64
51x+8y=193
-----------
43x = 129
x = 129/43
x = 3
.
To find y, substitute the above into:
x+ y= 8
3+ y= 8
y = 5
.
Our solution:
x = 3
y = 5

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A blue whale weighs approximately 100 tons, while a house cat weighs 5 pounds. About how many cats would it take to equal the weight of one blue whale?
.
A "ton" is equivalent to 2000 pounds.
Therefore, 100 tons is:
100*2000 = 200,000 pounds
.
200000/5 = 40,000 cats

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y = 6x – 15
is in the "slope-intercept" form
y = mx + b
where
m is slope
b is the y-intercept
.
To be parallel, slopes are identical.
So, our new line has
m = 6
and crosses
(0,0)
.
Plug the above into the "point-slope" form
y - y1 = m(x-x1)
y - 0 = 6(x-0)
y = 6x (this is what they're looking for)

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A bike shop rents mountain bikes for a $8.50 insurance charge plus $3.50 for each hour. For how many hours can a person rent a bike with $32.50.
.
Let h = number of hours rented
then
"rental cost" = "insurance" + "hourly charge"
32.50 = 8.50 + 3.50h
24 = 3.50h
6.86 hours = h
.

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Solve the system of equations.
Do this by the addition method -- add both equations together:
4x-y=6
x+y=4
-------
5x =10
x = 2
.
Substitute the above into:
x+y=4
2+y=4
y=2
.
Therefore, the graphs cross at:
(x,y) = (2,2)

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.
Factoring numerator and denominator:

.
Canceling like-terms:

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Apply Pythagorean theorem:
Let h = hypotenuse
then
h^2 = 3^2 + 5^2
h^2 = 9 + 25
h^2 = 34
h =

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A man mixes peanuts selling at 40 cents a pound with cashew nuts selling at 70 cents a pound to make a 50-pound mixture which he sells at 52 cents a pound. How many pounds of each kind of nuts must he mix?
.
Let x = pounds of peanuts
then
50-x = pounds of cashews
.
.40x + .70(50-x) = .52(50)
.40x + 35 -.70x = 26
35 -.30x = 26
35 = 26 + .30x
9 = .30x
30 pounds = x (peanuts)
.
Cashews:
50-x = 50-30 = 20 pounds (cashews)

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.
Setting each term to zero:
(3x-1)=0
3x = 1
x = 1/3
.
(x+4)=0
x = -4
.
Therefore,
x = {-4, 1/3}

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Tiger Woods hits a golf ball into the air. The equation that describs the path of the ball is h=55t-5t^2. There are three questions.
(A How high is it the ball after 2 seconds?)
h=55t-5t^2
Simply substitute 't' with 2 and solve:
h=55(2)-5(2)^2
h=110-5(4)
h=110-20
h=90 meters
.
(B When will it first reach 140 meters?)
h=55t-5t^2
Subsitute 'h' with 140 meters and solve:
140=55t-5t^2
0=55t-5t^2-140
0=-5t^2+55t-140
0=-t^2+11t-28
0=t^2-11t+28
0=(t-4)(t-7)
t = {4,7}
Ball reaches height at 4 and 7 seconds
.
(C When will it hit the ground?)
h=55t-5t^2
Subsitute 'h' with 0 meters and solve:
0=55t-5t^2
0=11t-t^2
0=-t^2+11t
0=t^2-11t
0=t(t-11)
t = {0,11}
Ball is at 0 meters at 0 seconds and 11 seconds

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LN(x+3)+LNx=LN4
LNx(x+3)=LN4
x(x+3)=4
x^2 + 3x = 4
x^2 + 3x - 4 = 0
(x-4)(x+1) = 0
x = {-1, 4}
We can toss out the negative solution (ln of a negative does not make sense), leaving us with:
x = 4

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3. The length of a rectangle is 16cm greater than its width. The area is 35m^2. Find the dimensions of the rectangle, to the nearest hundredth of a metre.
.
Let w = width
then
w+16 = length
.
w(w+16) = 35
w^2+16w = 35
w^2+16w-35 = 0
Using the quadratic equation, we get:
x = {1.95, -17.95}
We can toss out the negative solution leaving:
x = 1.95 m
.
Details of quadratic follows:

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Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=396 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 1.9498743710662, -17.9498743710662. Here's your graph:

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A 50 m by 120 m park consists of a rectangular lawn surrounded by a path of uniform width. Find the dimensions of the lawn if its area is the same as the area of the path. (Hint: Let x = the width of path)
.
Draw a diagram of the problem -- it'll help you see the problem.
.
Area of lawn:
(50-2x)(120-2x)
= 2(25-x)2(60-x)
= 4(25-x)(60-x)
= 4(150-85x+x^2)
= 4x^2 - 340x + 6000
.
Area of path:
(50)(120) - (50-2x)(120-2x)
= 6000 - (4x^2 - 340x + 6000)
= -4x^2 + 340x
.
Since:
"Area of lawn" = "Area of path"
we have
4x^2 - 340x + 6000 = -4x^2 + 340x
8x^2 - 340x + 6000 = 340x
8x^2 - 680x + 6000 = 0
x^2 - 85x + 750 = 0
(x-75)(x-10) = 0
x = {10, 75}
.
We can toss out the 75 leaving:
x = 10 meters (width of path)
.
dimensions of lawn:
width = 50 -2x = 50 - 20 = 30 meters
length = 120 -2x = 120 - 20 = 100 meters
Therefore, the lawn is
100 x 20 meters

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The LCD:
2(x+2)

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The sum of two numbers is 29. If the smaller is doubled and the larger is increased by 7, the resulting numbers will be equal. Find both numbers.
.
Let x = smaller number
and y = larger number
.
Since we have two unknowns, we need two equations:
From:"The sum of two numbers is 29" we get equation 1:
x + y = 29
.
From:"If the smaller is doubled and the larger is increased by 7, the resulting numbers will be equal." we get equation 2:
2x = y + 7
.
Using the "substitution method", solve equation 1 for y:
x + y = 29
y = 29 - x
.
Substitute the above into equation 2 and solve for x:
2x = y + 7
2x = 29 - x + 7
3x = 29 + 7
3x = 36
x = 12 (smaller number)
.
To find the larger number, substitute the above into equation 1 and solve for y:
x + y = 29
12 + y = 29
y = 29 - 12
y = 17 (larger number)

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Calculate temperature in Fahrenheit if the temperature in celsius is -60C
F=(9/5)C+32.
.
Simply plug and solve:
F=(9/5)C+32
F=(9/5)(-60)+32
F=(9)(-12)+32
F=-108+32
F=-76 (degrees Fahrenheit)

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y=-5x-1
.
The given equation is already in the "slope-intercept" form of a line:
y = mx + b
where
m is the slope
b is the y-intercept
.
Therefore, the slope is:
m = -5

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There are 220 barrels of cement in one shed and 510 barrels in another. How many barrels must be transferred from the second to the first shed will contain two thirds as much as the second?
.
Let x = number of barrels transferred
then
220 + x = (2/3)(510-x)
.
Now, solve for x:
220 + x = (2/3)(510-x)
Multiply both sides by 3:
660 + 3x = (2)(510-x)
660 + 3x = 1020 - 2x
660 + 5x = 1020
5x = 360
x = 72 barrels

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Looks just fine to me.
You should identify what 'x' is though:
I.E.
Let x = time it took to fly to Moscow

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Please help me find the root for sqrt y^36

=
=
.
Please help me simplify the expression sqrt 16n

=
=
.
Help me to use thee product rule to simplify sqrt 12x^8

=
=
.
Please help me to evaluate the expression for 25^-3/2

=
=
=
=
=
=
=

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Find two consecutive even integers x and x+2 with a product of 80. Write and equation and solve.
.
Equation:
x(x+2) = 80
.
Solving:
x^2+2x = 80
x^2+2x-80 = 0
(x+10)(x-8) = 0
x = {-10, 8}
.
Solutions:
8 and 10
-10 and -8

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5000=750(1.02)^(4t)
First, divide both sides by 750
6.6667 = (1.02)^(4t)
Next, take log base 1.02 of both sides:
log(6.6667)/log(1.02) = 4t
95.8014 = 4t
finally, dividing both sides by 4:
23.9504 = t

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Please help with the following question Write the equation for a line that is perpendicular to the line, y=3x+10/3 and goes thru the point [-15,6]
.
The slope of (from inspection)
y=3x+10/3
is 3
.
If a line is to be perpendicular the slope has to be the "negative reciprocal":
Let m = our new slope
then
3m = -1
m = -1/3 (our new slope)
.
Using our new slope and [-15,6]
plug it into the "point-slope" form:
y - y1 = m(x - x1)
y - 6 = (-1/3)(x - (-15))
y - 6 = (-1/3)(x + 15)
y - 6 = (-1/3)x - 5
y = (-1/3)x + 1 (this is what they're looking for)

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Since you are trying to add fractions, you must have the same denominator.
A "common denominator" is (z-4)(z+4)

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what is the area of an equilateral triangle with side lengths of 3?
.
Area of triangle = (1/2)bh
where
h is height
b is base (3)
.
Calculate height using Pythagorean theorem:
Let h = height
h^2 + (3/2)^2 = 3^2
h^2 + (9/4) = 9
h^2 = 9 - 9/4
h^2 = 36/4 - 9/4
h^2 = 27/4
h =
.
Plugging it all back into:
Area of triangle = (1/2)bh
Area of triangle =
Area of triangle =

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72
= 2*36
= 2*2*18
= 2*2*2*9
= 2*2*2*3*3 (this is what they're looking for)

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Slope, given two point is:
(y2 - y1)/(x2 - x1)
.
For:
(-3,7), and (5,-5)
We plug it in to:
(y2 - y1)/(x2 - x1)
= (-5 - 7)/(5 - (-3))
= (-12)/(5+3)
= (-12)/8
= -3/2 (this is your slope)

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What is the length of a diagonal of a rectangle with a length of 8meters and a width of 6 meters?
.
Apply Pythagorean theorem:
Let d = diagonal
then
d^2 = 8^2 + 6^2
d^2 = 64 + 36
d^2 = 100
d = 10 meters (length of diagonal)