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log(base3)(2x-1)=3
2x-1=3^3
2x-1=27
2x=28
x=14

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y^2+y-12
= (y+4)(y-3)

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y^2+12y-13
= (y+13)(y-1)

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log3x+log(x+2)=1
log3x(x+2)=1
3x(x+2)=10^1
3x^2+6x=10
3x^2+6x-10=0
Solving the above using the quadratic equation yields two solutions:
x = {1.082, -3.082}
.
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=156 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 1.08166599946613, -3.08166599946613. Here's your graph:

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log(x+9)-logx=3
log(x+9)/x=3
(x+9)/x=10^3
(x+9)/x=1000
(x+9)=1000x
9=999x
1/111 = x

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3log2x=4
log(2x)^3=4
(2x)^3 = 10^4
8x^3 = 10000
x^3 = 1250
x = 10.772

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log(3x+7)=1
(3x+7)= 10^1
3x+7 = 10
3x = 3
x = 1

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2log4-log3+logx-4=0
log4^2-log3+logx-4=0
log(4^2/3)+logx-4=0
log(16/3)+logx-4=0
log(16(x-4)/3)=0
16(x-4)/3 = 10^0
16(x-4)/3 = 1
16(x-4) = 3
x-4 = 3/16
x = 3/16 + 4
x = 3/16 + 64/16
x = 67/16

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5Ln(1-x)= 8
Ln(1-x)^5= 8
(1-x)^5= e^8
1-x = fifthRoot(e^8)
1 = fifthRoot(e^8)+x
1 - fifthRoot(e^8) = x
1 - fifthRoot(2980.958) = x
1 - 4.953 = x
-3.953 = x

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Factoring numerator and denominator:

.
Canceling terms, we get:

.
Or, we could rewrite it as:

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You can review this site:
http://www.regentsprep.org/Regents/math/algtrig/ATP2/GeoSeq.htm
.
An = A1(r^(n-1))
.
The idea is that because they gave you:
a3 = 32 and a5 = 512
you can now find A1(first term) and r (the common ratio).
Now, you can find any term after that.
.
An = A1(r^(n-1))
32 = A1(r^2)
A1 = 32/r^2
.
An = A1(r^(n-1))
An = A1(r^4)
512 = (32/r^2)(r^4)
512 = 32(r^2)
16 = r^2
4 = r (common ratio)
.
A1 = 32/r^2 = 32/4^2 = 32/16 = 2 (first term)
.
An = A1(r^(n-1))
a8 = (2)(4^(8-1))
a8 = (2)(4^7)
a8 = 2*16384
a8 = 32768

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A farmer has 336 running feet of fence and wishes to form a rectangular pasture. One side of the pasture will be bounded by a very long wall, and so no fencing material will be needed for that side of the pasture. What should the dimensions of the pasture be if the area of the pasture is to be largest possible?
.
Let x = width
and y = length
.
From the provided length of fencing available:
2x+y = 336
Solving for y:
y = 336-2x
.
Area = xy
Plugging in our value of y:
Area = x(336-2x)
Area = 336x-2x^2
Area = -2x^2 + 336x (parabola that opens downwards)
.
If we find the vertex, we'll find the maximum value of the width:
Area = -2x^2 + 336x
Area = -2(x^2 - 168x)
Area = -2(x^2 - 168x + 7056) + 14112
Area = -2(x - 84)^2 + 14112
Since the vertex is at (84, 14112)
we know the width is 84 feet
.
Length:
2x+y = 336
2(84)+y = 336
168+y = 336
y = 168 feet (length)
.
Dimension:
84 feet by 168 feet

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The "vertex form" is
y= a(x-h)2+k
where
(h,k) is the vertex
.
The idea is to make your equation look like that:
f(x) = (x - 1)^2 + (x + 3)
f(x) = (x - 1)(x - 1) + (x + 3)
f(x) = x^2 - 2x + 1 + x + 3
f(x) = x^2 - x + 4
Group x-terms:
f(x) = (x^2 - x) + 4
Complete the square:
f(x) = (x^2 - x + __ ) + 4
f(x) = (x^2 - x + 1/4 ) + 4 -1/4
f(x) = (x - 1/2)^2 + 16/4 -1/4
f(x) = (x - 1/2)^2 + 15/4
.
vertex = (1/2, 15/4)

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The "vertex form" of a parabola is:
y= a(x-h)^2+k
where
(h,k) is the vertex
.
So, the idea is to manipulate the original equation into the "vertex form":
f(x) = -3x^2 + 5x + 1
First, group the x terms:
f(x) = (-3x^2 + 5x) + 1
Factor out a -3:
f(x) = -3(x^2 + (-5/3)x) + 1
Complete the square:
f(x) = -3(x^2 + (-5/3)x + __ ) + 1
f(x) = -3(x^2 + (-5/3)x + (25/36) ) + 1 + 25/12
f(x) = -3(x - 5/6 )^2 + 12/12 + 25/12
f(x) = -3(x - 5/6 )^2 + 37/12 (this is the vertex form)
.
Vertex = (5/6, 37/12)

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Two angles atre complementary. The sum of the measure of the first angle and one-fourth the second angle is 69 degrees. Find the measure of the angles.
.
Two angles are complementary if their sum equals 180 degrees.
.
Let x = measure of second angle
then
180-x = measure of first angle
.
(180-x) + (1/4)x = 69
720 - 4x + x = 276
720 - 3x = 276
-3x = -444
x = 148 degrees (second angle)
.
First angle:
180-x = 180-148 = 32 degrees (first angle)

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There were twice as many dimes as nickles and the value of the dimes was $.90 more than the value of the nickles.
.
Let n = number of nickels
then
2n = number of dimes
.
.10(2n) = .05n + .90
.20n = .05n + .90
.15n = .90
n = .90/.15
n = 6 (nickels)
.
Dimes:
2n = 2(6) = 12 (dimes)

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x^4 + 5x^2 -36 = 0
Factoring:
(x^2+9)(x^2-4) = 0
.
Setting each term to zero:
(x^2+9) = 0
x^2 = -9
x = sqrt(-9)
x = +- 3i
.
(x^2-4) = 0
x^2 = 4
x = sqrt(4)
x = +- 2

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To solve a linear equation, isolate the unknown to one side of the equat
ion.
.
12x + 6 = 31 – 2x
.
Start by moving all the x terms to the left. Do this by ADDING 2x to both sides:
12x + 6 + 2x = 31 – 2x + 2x
14x + 6 = 31
.
Next, subtract 6 from both sides:
14x + 6 - 6 = 31 - 6
14x = 25
.
Finally, MULTIPLY both sides by (1/14):
(1/14)14x = 25(1/14)
x = 25/14 (solution for x)

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Manipulate the equation into the "slope-intercept" form:
y = mx + b
where
m is slope
b is the y-intercept
.
3x-5y+5=0
3x-5y= -5
-5y= -3x -5
5y= 3x + 5
y = (3/5)x + (5/5)
y = (3/5)x + 1
.
By inspection, based on the "slope-intercept" form we have:
slope is 3/5
y-intercept is at (0,1)

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(-8c^2+4c+2)-(6c^2+2)
.
Distribute the negative to terms inside the right parenthesis:
-8c^2+4c+2-6c^2-2
.
Group like-terms:
(-8c^2-6c^2)+4c+(2-2)
.
Combine:
(-14c^2)+4c+(0)
.
-14c^2+4c
Or, factoring out 2c:
2c(-7c + 2)

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The area of a rectangle is 64cm^2. Its length is 4times the width. What is its length and width
.
Let w = width
then
4w = length
.
w(4w) = 64
4w^2 =64
w^2 = 16
w = 4 cm (width)
.
Length:
4w = 4(4) = 16 cm (length)

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(2x^2-3xy+y^2)+(-8x^2-8xy-y^2)+(x^2+xy-4Y^2)
.
Remove parentheses:
2x^2-3xy+y^2-8x^2-8xy-y^2+x^2+xy-4Y^2
.
Group like-terms:
(2x^2-8x^2+x^2)+(-3xy-8xy+xy)+(y^2-y^2-4Y^2)
.
Combine like-terms:
(-5x^2)+(-10xy)+(-4Y^2)
-5x^2-10xy-4Y^2
or
-(5x^2+10xy+4Y^2)

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Isolate "b-a" to one side of the equation.
.
If you started with:
13+a=25+b
Subtract 'a' from both sides:
13 = 25+b-a
Now, subtract 25 from both sides:
13-25 = b-a
-12 = b-a (this is what they're looking for)

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If jan can paint a house in 4 hours and jim can paint a house in 6 hours, how long will it take for them to paint the house together.
.
Let x = hours for both to paint house
then
x(1/4 + 1/6) = 1
x/4 + x/6) = 1
6x + 4x = 24
10x = 24
x = 2.4 hours
or
2 hours and 24 minutes

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Find the slope -intercept equation of the line that has the given characteristics
slope 3 and y -intercept (0,9)
.
The "slope-intercept" form of a line is
y = mx + b
where
m is slope
b is the y-intercept
.
Therefore, for:
slope 3 and y -intercept (0,9)
we have
y = 3x + 9 (this is what they're looking for)

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For it to be a function, for any value of 'x', there should only be one value of 'y',.
.
The ONLY answer that fits would be: A.

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how can the number 45 be divided into two parts so that 4 times one part is 9 less than 5 times the other?
.
Let x = first part
then
45-x = second part
.
4(45-x) = 5x-9
180 - 4x = 5x-9
180 = 9x-9
189 = 9x
21 = x (first part)
.
Second part:
45-x = 45-21 = 24 (second part)

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The area of a triangle is 30 square inches. The base of the triangle measures 2 inches more than twice the height of a triangle. Find the measures of the base and the height.
.
Area of any triangle = (1/2)bh
where
b is base
h is height
.
Let h = height
then
2h+2 = base
.
30 = (1/2)(2h+2)h
30 = (h+1)h
30 = h^2+h
0 = h^2+h-30
0 = (h+6)(h-5)
.
h ={-6,5}
We can toss out the negative solution leaving us with:
h = 5 inches
.
base:
2h+2 = 2(5)+2 = 10+2 = 12 inches
.
Therfore,
height is 5 inches
base is 12 inches

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x^2=6x-3
Move all terms to the left:
x^2-6x+3 = 0
Since we can't factor, we must resort to the quadratic equation.
Doing so, yields two solutions:
x = {5.449, 0.551}
.
Details of quadratic equation 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=24 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 5.44948974278318, 0.550510257216822. Here's your graph:

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5(-2x+1)= -9(x+1)
Distributing on the left and right side:
-10x+5= -9x-9
5= x-9
14 = x

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Hi there, I need some help again. Can someone help me solve this? I don't even understand the problem. Thanks! The perimeter of a rectangle is 108m. The length is 9m more then twice the width find the dimensions. What is the length.
.
Let w = width of rectangle
then
2w+9 = length of rectangle
.
2(width + length) = perimeter
.
2(w + 2w+9) = 108
2(3w+9) = 108
6w+18 = 108
6w = 90
w = 16 m
.
length:
2w+9 = 2(16)+9 = 32+9 = 41 m