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Your gas tank was full when you left the town. Now it needs 16 gallons to fill up. That means you used 16 gallons during the trip. You also drove for 528 miles during that time.
Gas mileage = miles / gas used
= 528 / 16
= 33
Your gas mileage is 33 miles per gallon.

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let x be the first number
let 54-x be the second number
(second statement)
x - 6 = 2(54-x)
x - 6 = 108 - 2x
3x = 114
x = 38
54-x=16
The first number is 38 and the second number is 16.

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So you have no numbers. I'm assuming you're using variables then.
Let x be the yearly interest rate.
Let $y be the investment after 1 year.
This is the equation for simple interest for 1 year.
Original Investment*(1+x)=y
So if you have y and want to know the original investment,
Original Investment = y/(1+x)

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|x-2|=3x+1
The first stop of finding absolute value functions is by finding the breakpoints. This is when the function changes in normal terms. The break point is basically just the zero of the absolute function.
x-2=0
x=2
The breakpoint is at 2. So when x is greater than 2, |x-2| will behave just like x-2. But, when x-2 is less than 2, |x-2| will behave like -(x-2). SO, to the solving part.
CASE 1: (Assume x is greater or equal to 2)
x-2 = 3x + 1
-3 = 2x
x = -3/2
This doesn't follow with our assumption. We can conclude that x isn't >= 2.
CASE 2: (Assume x is less than 2)
-(x-2) = 3x + 1
-x + 2 = 3x + 1
1 = 4x
x = 1/4
This agrees with our assumption.
After studying both cases we can conclude that x is equal to 1/4.

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f(x) = 4x^2/x^2-9
I'm assuming you're asking for Vertical and Horizontal asymptote.
The Vertical Asymptote occurs when the denominator is equal to zero.
x^2-9=0
(x+3)(x-3)=0
x= 3, -3
So the vertical Asymptotes occur at x=3 and x=-3.
To find the asymptote the method I learned was to divide the top and bottom by 1/x^2. You divide top and bottom by 1/x^n(highest power in numerator).This is basically multiplying by 1 but it helps you out.
(4x^2/x^2)/(x^2/x^2 - 9/x^2)
=4/(1-9/x^2)
Now we take the limit of the as x-> infinity and negative infinity. This will mean that 9/x^2 becomes practically zero.
so we get 4/1 or 4. The graph will approach 4 when the x values approach positive or negative infinity.
This means the horizontal Asymptote is at y=4.

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4^y+4^(2*y+1)=1
Simplify the question by making a substitution.
Let x = 4^y

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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=17 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 0.390388203202208, -0.640388203202208. Here's your graph:

0.390388203202208, -0.640388203202208
sub into x = 4^y
logx = ylog4
y = logx/log4
y1 = -0.6785
y2 = invalid (can't log negative numbers)
The answer is approximately -0.6785. If you want exact formula you'll need to use the quadratic formula to solve and leave in exact numbers.

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(x^2 - 6x + 9) / (x^2 - 9)
factor
=(x-3)(x-3)/(x-3)(x+3)
FYI a^2 - b^2 = (a+b)(a-b)
cancel out (x-3)
=(x-3)/(x+3) I think that is fully simplified. You might also have to say that x can not equal positive or negative 3.

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Okay for elimination we need to eliminate something. I'll choose to eliminate x. I need both x's to be equal. So I'll multiply the first equation by 5 and the second equation by 3.
3x-5y=30 ... (i)
15x - 25y = 150 ... (iii)
5x + 7y = 4 ... (ii)
15x + 21y = 12 ... (iv)
Now eliminate. I will subtract equation 4 from equation 3.
15x - 25y = 150
-)15x+21y = 12
________________
0x -46y = 138
y = -138/46
y = 3
now I will sub y = 3 into equation 1.
3x - 5(3) = 30
3x - 15 = 30
3x = 45
x = 45/3
x = 15
Therefore, x = 15, y = 3. This was solved via elimination method.

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I'm assuming you mean solve for x in terms of y and w.
y = w/x - 2
Add two on both sides
y + 2 = w / x
Multiply both sides by x
x ( y + 2 ) = w
Divide both sides by y + 2
x = w / (y + 2)

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Lets say the original pool is l*w*h.
New pool is 2l*2w*h
a) the SA of bottom is the length times width.
Original Pool = l*w
New pool = 2l*2w = 4*l*w
FALSE
b) 4 walls is length times base times two. Added with width times base times two.
ORiginal Pool = l*b*2 + w*b*2
New Pool = 2l*b*2 + 2w*b*2 = 2(l*b*2 + w*b*2)
TRUE
c) Area of pool is length times width times base.
Original Pool = l*w*b
New Pool = 2l*2w*b = 4*l*w*b
FALSE
The correct answer is b)

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The expression is undefined if the bottom of denominator is 0 as you can't /0.
So, we want to see when the bottom is 0.
...




or
or
Therefore, if m is equal to 1 or -3/2 than the expression is undefined.

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Let x m be the width of the table top.
Let x+5 m be the length of the table top.
...
A=l*w
66=(x)(x+5)
66=x^2+5x
0=x^2+5x-66

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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=289 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 6, -11. Here's your graph:

The answer must be positive so,
x=6
x+5=11
...
Therefore, the dimensions of the table top is 6m by 11m.

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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=25 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 5, 0. Here's your graph:

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log (a^1/3 b^1/2) /c
= log (a^1/3 b^1/2) - log c
= log a^1/3 + log b^1/2 - log c
= 1/3*log a + 1/2*log b - log c

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x^2+14x+40=-5
x^2+14x+40+5=0
x^2+14x+45=0

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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=16 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: -5, -9. Here's your graph:

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Let x be the number of games they won.
Let x-12 be the number of games they lost.
...
x+x-12=52
2x-12=52
2x=52+12
2x=64
x=32
Therefore, the Tigers won 32 games that season.

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You mean log root x
logx 216 = 3
x^3 = 216
x = +- 6

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n(n+5)=-4
n^2+5n=-4
n^2+5n+4=0

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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=9 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: -1, -4. Here's your graph:

Therefore n = -1 or -4.
3n^2= 3 or 48

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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=1 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: -1, -2. Here's your graph:

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Let x ft be the width of the rectangle.
Let 3x-1 ft be the length of the rectangle.
...
l*w=A
(3x-1)(x)=30
3x^2-x=30
3x^2-x-30=0

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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=361 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 3.33333333333333, -3. Here's your graph:

x must be positive.
x=3.33
3x-1=9
Therefore, the rectangle is 3.33 ft by 9 ft.

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Let x m be the width of the rectangle.
Let 8x m be the length of the rectangle.
...
P=2(l+w)
200=2(8x+x)
100=9x
x=100/9
x=11.1
8x=88.9
Therefore the dimensions of the rectangle are 11.1 m by 88.9 m.

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... difference of squares
or
or

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Let x ft be the width of the garden.
Let x+7 ft be the length of the garden.
...
... Pythagorean theorem

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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=289 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 5, -12. Here's your graph:

x must be positive.
x=5
x+7=12
Therefore, the dimensions of the garden is 5 ft by 12 ft.

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A=b*h
A=12*6
A=72
Therefore the area of the parallelogram is 72 inches squared.

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1n (x+7) + 1n (x+3) = 1n77
ln x + ln 7 + ln x + ln 3 = ln 77 ****** Problem here
---------------------
2 ln x + ln 7*3 = ln77
2 ln x = ln 77 - ln 21
ln x^2 = ln 77/21
x^2 = 77/21
x = sqrt (77/21)
x = 1.915
x must be positive as you can't have negative ln.

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log 3=2 with log base x-1

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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=12 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 2.73205080756888, -0.732050807568877. Here's your graph:

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Division of complex numbers is

if you need more info or proof search complex numbers on Wikipedia.
...

=
=

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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=441 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: -1, 1.625. Here's your graph:

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Convert them both to the same base.
...



Therefore, since, therefore gloop=junk.

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Let x be the first integer.
Let x+1 be the second integer.
Let x+2 be the third integer.
...
(x)(x+1)=x+2+62
x^2+x=x+2+62
x^2+x-x-64=0
x^2-64=0
x^2-(8)^2=0
(x+8)(x-8)=0
x=-8 or x=8
x+2= -6 or 10
...
x=10 as that is the the forth choice d.

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Let x be the smaller number.
Let x+4 be the larger number.
...
(x+4)+5=4x
x+4+5=4x
9=4x-x
9=3x
x=9/3
x=3
x+4=7
Therefore, the two numbers are 3 and 7.