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Two trains 100 km apart each travel towards each other at 10 km per hour. A bumblebee starts flying back and forth between the two trains at 60 km per hour, until the two trains crash into each other. How far does the bumblebee travel?
.
You will need to apply the "distance formula":
d = rt
where
d is distance
r is rate or speed
t is time
.
Let t = time before trains crash
.
10t + 10t = 100
20t = 100
t = 5 hours
.
Since the bee is traveling 60 km/hr we multiply this with 5 hours to determine distance traveled:
60*5 = 300 km

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.
Let w = width of aluminum
then
2w = length of aluminum
.
Box AFTER the square is cut:
width = w-5-5 = w-10
length = 2w-5-5 = 2w-10
.
Volume of box:
10500 = (w-10)(2w-10)5
2100 = (w-10)(2w-10)
2100 = (w-10)2(w-5)
1050 = (w-10)(w-5)
1050 = w^2-5w-10w+50
1050 = w^2-15w+50
0 = w^2-15w-1000
.
Factoring the right:
0 = (w+25)(w-40)
.
w = {40, -25}
.
We can toss out the negative solution leaving us with:
w = 40 cm (width of rectangle)
.
length of rectangle:
2w = 2(40) = 80 cm (length of rectangle)

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3x^2-2x+1=0
.
Factoring the left side:
(3x+1)(x-1) = 0
.
Setting each term on the left to zero to find the set:
3x+1 = 0
3x = -1
x = -1/3
.
x-1 = 0
x = 1
.
Solution set:
x = {-1/3, 1}

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Group terms in the denominator:



Factoring the numerator:



Applying factor for "difference of cubes" in the denominator:

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Start by moving all the terms to one side:
5z^2 = 2z + 3
Subtracting both sides by 2z+3:
5z^2 - 2z - 3 = 0
.
Always attempt to factor first, then failing that use the quadratic equation.
.
In this case, you must use the quadratic equation. Doing so will yield:
z = {1, -0.6}
.
Details of quadratic below:

Solved by pluggable solver: SOLVE quadratic equation with variable

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=64 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 1, -0.6. Here's your graph:

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This site gives you a very good tutorial on the subject:
http://www.purplemath.com/modules/systlin4.htm

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4m^2 - 8m + 1 = 0
.
To solve,
always try factoring first
failing that, use the quadratic equation.
.
In this case, we must use the quadratic equation.
Doing so will yield:
m = {1.866, 0.134}
.
Details below:

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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=48 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 1.86602540378444, 0.133974596215561. Here's your graph:

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Mixture_Word_Problems/169880 (2008-11-27 06:59:42): Tony has two kind of tea, one selling at P280 per kilogram and the other at P305 per kilogram. How many kilograms of each should she get from to make a 25-kilogram blend which will sell at P290 per kilo?
.
Let x = amount of P280 per kilogram
then
25-x = amount of P305 per kilogram
.
280x + 305(25-x) = 290(25)
280x + 7625 - 305x = 7250
375 - 25x = 0
375 = 25x
15 kilograms = x (amount of P280 per kilogram)
.
amount of P305 per kilogram:
25-x = 25-15 = 10 kilograms

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C(p)= 500000/(100-p)
.
a) Use the accompanying graph to estimate the cost for
removing 90% and 95% of the toxic chemicals.
At 90%
C(90)= 500000/(100-90)
C(90)= 500000/10
C(90)= $50,000
.
At 95%
C(95)= 500000/(100-95)
C(95)= 500000/5
C(95)= $100,000
.
b) Use the formula to find C(99.5) and C(99.9).
At 99.5%
C(99.5)= 500000/(100-99.5)
C(99.5)= 500000/.5
C(99.5)= $1,000,000
.
At 99.9%
C(99.9)= 500000/(100-99.9)
C(99.9)= 500000/0.1
C(99.9)= $5,000,000
.
c) What happens to the cost as the percentage of
pollutants removed approaches 100%?
As it approaches 100%, the denominator goes to zero -- causing C(p) to become undefined -- or, reach an infinitely high cost.

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Reference:
http://www.mathwarehouse.com/geometry/parabola/standard-and-vertex-form.php
.
The vertex form of a parabola:
y= a(x-h)^2+k
.
Where
(h,k) is the vertex
.
Here they give you the vertex as:
(h,k) = (-1,2)
.
y= a(x-(-1))^2+2
And since the problem stated that is was a MINIMUM... 'a' is POSITIVE.
y= (x+1)^2+2
y= x^2+2x+1+2
y= x^2+2x+3

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HARRY IS FIVE YEARS OLDER THAN DON. THREE YEARS AGO DON WAS TWO-THIRDS AS OLD AS HARRY. HOW OLD IS HARRY?
.
Let x = Don's age
then from "HARRY IS FIVE YEARS OLDER THAN DON." we have:
x+5 = Harry's age
.
Finally, from "THREE YEARS AGO DON WAS TWO-THIRDS AS OLD AS HARRY." we get our equation: (three years ago means current year minus 3)
x-3 = (2/3)(x+5-3)
x-3 = (2/3)(x+2)
3x-9 = 2(x+2)
3x-9 = 2x+4
x-9 = 4
x = 13 years (Don's age)
.
Harry's age:
x+5 = 13+5 = 18 years

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A rectangular lap pool is filled with water to a depth of 5 feet. If the pool is 6 feet wide and 200 feet long, how many gallons would be required, if each cubic foot equals approximately 7.5 gallons?
.
Volume of pool (filled to a depth of 5 ft) is:
6*200*5 = 3600 cubic feet
.
If each cubic feet equals 7.5 gallons:
3600 * 7.5 = 27,000 gallons

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Calculate the volume of a sphere with a diameter of 10 inches. Round answer to the nearest cubic inch.
.
Volume of a sphere = (4/3)(pi)r^3
where
pi is 3.14
r is radius
.
radius = (1/2)diameter = (1/2)10 = 5 inches
.
Volume of a sphere = (4/3)(pi)r^3
Volume of a sphere = (4/3)(3.14)5^3
Volume of a sphere = (4/3)(3.14)125
Volume of a sphere = (166.667)(3.14)
Volume of a sphere = 523.3333
Volume of a sphere = 523 cubic inches

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The perimeter of a swimming pool is 96 meters. The width is 3 meters more that the length of the pool. Find the length and width of the pool.
.
Let x = length of the pool
then
x+3 = width of the pool
.
Remember, the perimeter of any rectangle is:
W + W + L + L
2W + 2L
2(W + L)
.
So, rewriting the above with our variable we have:
96 = 2(x + x+3)
96 = 2(2x+3)
96 = 4x+6
90 = 4x
22.5 meters = x (length of pool)
.
width:
x+3 = 22.5+3 = 25.5 meters (width of pool)

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The perimeter of a rectangle window is 32 ft. Find the dimensions of the window that will enclose the largest area.
.
Let W = width
and L = length
.
Since perimeter is:
2(L + W) = 32
L + W = 16
L = 16-W
.
Area = W(16-W)
Area = -W^2 + 16W
.
By inspection, we see (from the -1 coefficient associated with W^2) that it is a parabola which opens downward -- therefore, the "axis of symmetry" should give you the max.
.
Axis of symmetry:
W = -b/2a
W = -16/2(-1)
W = 8 feet (width)
.
Length:
16-W = 16-8 = 8 feet (length)

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A photo is 3 inches longer that it is wide. A 2-inch border is placed around the photo making the total area of the photo and border 108 square inches. What are the dimensions of the photo?
.
Let w = width of photo
then
w+3 = length of photo
.
If a 2-inch border is added:
width is:
w+2+2 = w+4
.
length is:
w+3+2+2 = w+7
.
(w+4)(w+7) = 108
w^2 + 7w + 4w + 28 = 108
w^2 + 11w + 28 = 108
w^2 + 11w - 80 = 0
.
Factoring we get:
(w+16)(w-5) = 0
w = {5, -16}
.
We can toss out the negative solution so we get:
w = 5 inches (width)
.
Length:
w+3 = 8+3 = 11 inches (length)

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Francine is earning $7.00 per hour. She gets a 20% raise in pay. By what fraction should you multiply to find how much more per hour she will now make? What will Francine's new hourly rate be?
~~READ~~ the problem and break it into parts to solve it. Identify any irrelevant information in the problem.
No irrelevant information.
~~SHOW/WRITE
% is "per one hundred"
therefore
20% in fractional form is 20/100
.
So, if Francine got a 20% raise it would be:
7 * 20/100
= 7 * .20
= $1.40 (raise)
.
~~WRITE~~ your answer.
New hourly rate then is:
7 + 1.40 = $8.40 per hour

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1/x - 1/(x-1)= -1/6
.
Start by multiplying both sides with a common denominator: 6(x)(x-1)
.
6(x)(x-1)[1/x - 1/(x-1)]= 6(x)(x-1)[-1/6]
6(x-1) - 6(x) = -(x)(x-1)
6x - 6 - 6x = -x^2 + 1
- 6 = -x^2 + 1
- 6 + x^2 = 1
x^2 = 7
x = (+-)sqrt(7)

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2P^2-2P-5=0
.
Always try to factor first, failing that you can always use the "quadratic equation". In this case, you can't factor so you must use the quadratic equation.
.
Doing so will yield:
P = {2.158, -1.158}
.
Details of the quadratic below:

Solved by pluggable solver: SOLVE quadratic equation with variable

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=44 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 2.1583123951777, -1.1583123951777. Here's your graph:

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Applying "log rules":
log9(x-7) + log9(x-7) = 1
log9((x-7)(x-7)) = 1
(x-7)(x-7) = 9^1
(x-7)(x-7) = 9
x^2-7x-7x+49 = 9
x^2-14x+49 = 9
x^2-14x+40 = 0
(x-10)(x-4) = 0
.
x = {10,4}

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You forgot the "r!" in the denominator.
C(n,r) = n!/(r!(n-r)!)
C(12,10) = 12!/(10!(12-10)!)
C(12,10) = 12!/(10!(2)!)
C(12,10) = (11*12)/(1*2)
C(12,10) = 132/2
C(12,10) = 66

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Divide 556 into two parts such that if the larger part is added to 12 and the smaller part is added to 18 the resulting sums would be equal.
.
Let L = larger part
and S = smaller part
.
Since we have two unknowns, we'll need two equations:
L + S = 556 (equation 1)
L + 12 = S + 18 (equation 2)
.
Solving equation 1 for S:
L + S = 556
S = 556 - L
.
Plug the above into equation 2 and solve for L:
L + 12 = S + 18
L + 12 = 556 - L + 18
L + 12 = 574 - L
2L + 12 = 574
2L = 562
L = 281 (larger number)
.
To find the smaller number, plug the above into equation 1:
L + S = 556
281 + S = 556
S = 556 - 281
S = 275 (smaller number)

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Sure, what is your equation?

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(1-2x)(3x+1)-(x-1)^2
What I have done is this:
(x-1)^2 = x^2 -2x +1
And using FOIL, (1-2x)(3x+1) =
3x +1 -6x^2 -2x
So far, so good!!
.
Now, rewriting:
(1-2x)(3x+1)-(x-1)^2
We get:
(3x +1 -6x^2 -2x) - (x^2 -2x +1)
3x +1 -6x^2 -2x - x^2 + 2x - 1
3x -6x^2 -2x - x^2 + 2x
3x -6x^2- x^2
3x -7x^2
factoring out an 'x':
x(3-7x) (Which is answer 'a')

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a.
Should be:
p = -x + 62
price
*************************************************
b. YES!
R = xp
R = x(-x + 62)
R = -x^2 + 62x
revenue
*************************************************
c. YES
b=300
*************************************************
d. YES
C=6x+300
*************************************************
e. NO -- you have a "sign" problem.
profit = "revenue" - "cost"
P(x) = (-x^2 + 62x) - (6x+300)
P(x) = -x^2 + 62x - 6x - 300
P(x) = -x^2 + 56x - 300
*************************************************
f. Since you had e. wrong, this is wrong
P(x) = -x^2 + 56x - 300
P(20) = -(20^2) + 56(20) - 300
P(20) = -(400) + 1120 - 300
P(20) = -700 + 1120
P(20) = $420
*************************************************
g. Since you had e. wrong, this is wrong
P(x) = -x^2 + 56x - 300
P(25) = -(25^2) + 56(25) - 300
P(25) = -(625) + 1400 - 300
P(25) = -925 + 1400
P(25) = $475
*************************************************
h. Since you had e. wrong, this is wrong
P(x) = -x^2 + 56x - 300
P(0) = -0^2 + 56(0) - 300
P(0) = -300
Even if you don't sell anything, you STILL have costs -- rent, insurance, and wages, etc.
*************************************************
i.
Since the "profit" is increasing in part f. and g.
Try a larger number than 25 such as 35
P(x) = -x^2 + 56x - 300
P(35) = -(35^2) + 56(35) - 300
P(35) = -(1225) + 1960 - 300
P(35) = 435
Since it went DOWN from g. -- it's between 25 and 35...
The idea is to keep trying until you find the most profit...

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Let x = amount of 12% acid added
.
"amt of acid from 23%" + "amt of acid from 12%" = "amt in final mixture"
.23(440) + .12x = .145(440+x)
101.2 + .12x = 63.8 + .145x
37.4 + .12x = .145x
37.4 = 0.025x
1496 liters = x

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They give you:
x = 35
.
Plug it into the given equation and solve:
H(x)=2.75x+71.48
H(35)=2.75(35)+71.48
H(35)=96.25+71.48
H(35)=167.73 cm

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We can write "logbase2 4" as "log2(4)".
.
You will need to apply "log rules". You can review them at:
http://www.purplemath.com/modules/logrules.htm
.
log2(4) + log2(27)= log2(x)
log2(4*27)= log2(x)
log2(108)= log2(x)
108 = x

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You're welcome!
.
Basically, it's:
"Cost" = "cost of tile sets" + "fixed costs"
.
The problem gives you the "fixed costs" as $300
and
the "cost of tiles sets" is $6 for each "tile set"
So, if 'x' is the number of "tile sets" it would be:
6x
.
Pulling it all together:
C = 6x + 300 (this is what they're looking for)

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Suppose that a market research company finds that at a price of p = $20, they would sell x = 42 tiles each month. If they lower the price to p = $10, then more people would purchase the tile, and they can expect to sell x = 52 tiles in a month’s time. Find the equation of the line for the demand equation. Write your answer in the form p = mx + b. (Hint: Write an equation using two points in the form (x,p)).
.
Your "profit" formula is:
P(x)
.
The two (x,p) points they give you are:
(x,p) = (52,10)
(x,p) = (42,20)
.
The "form" they want it in is:
p = mx + b
where
p is the profit
m is slope
x is the tiles sold
b is the "y-intercept"
.
To find 'm', the slope (given two points):
m = (p2-p1)/(x2-x1)
plugging in your points:
m = (20-10)/(42-52)
m = 10/-10
m = -1
.
Now, using any one of the points (let's use 42,20) and the slope plug it back into (to find 'b'):
p = mx + b
20 = (-1)(42) + b
20 = -42 + b
62 = b
.
Recapping, we have:
m = -1
b = 62
.
To complete, insert the above into:
p = mx + b
p = (-1)x + (62)
p = -x + 62 (This is what they're looking for)

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You will need to apply "log rules". For a refresher see:
http://www.purplemath.com/modules/logrules.htm
.
log5x=log4+log(x+1)
log5x=log(4(x+1))
log5x=log(4x+4)
5x=4x+4
x=4