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2 - |x| ≥ -2
-|x| ≥ -4
|x| ≤ 4
x ≤ 4 and x ≥ -4
4 ≥ x ≥ -4

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This is actually a geometric sum or series.
S = a1(1 - r^n)/(1 - r)
S: sum, a1: first term, r: rate, n: # of terms
A man has a job for thirty days, he will be paid 1 penny for the first day, 2 pennies for the second day, 4 pennies for the third day, and so on. What is the equation that would give the total amount of pennies for the 30th day?
1 + 2 + 4 + 8 + ... a1 * r^(n - 1) or 2^(n - 1)
a1 = 1, r = 2, n = 30
S = (1 - 2^30)/(1 - 2) = 1,073,741,823 pennies or $10,737,418.23

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ln( y - 3 ) = ln( 2x^2 ) + ln( C )
ln( y - 3 ) = ln( 2C*x^2 )
y - 3 = 2C*x^2
y = 2C*x^2 + 3

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cos^2(3x) - sin^2(3x) = cos(6x)
cos^2(3x) - sin^2(3x) = cos(3x + 3x)
cos^2(3x) - sin^2(3x) = cos(3x)cos(3x) - sin(3x)sin(3x)
~
sin^2(2α) / sin^2(α) = 4 - 4sin^2(α)
sin(2α)sin(2α) / sin^2(α) = 4 - 4sin^2(α)
2sin(α)cos(α)2sin(α)cos(α) / sin^2(α) = 4 - 4sin^2(α)
4cos^2(α) = 4 - 4sin^2(α)
4(1 - sin^2(α)) = 4 - 4sin^2(α)

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When rotating a conic section, combinations of sin? and cos? are often used ~ yes, each is used
When rotating a conic section, only sin? needs to be used ~ no, cosine is used as well
Conic sections can neither be rotated nor translated ~ no, they can be rotated and moved
Conic sections can be rotated but not translated ~ no, they can be rotated and moved

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cos^4(x) - sin^4(x) = cos(2x)
cos^4(x) - sin^4(x) = 2cos^2(x) - 1
( cos^2(x) + sin^2(x) )( cos^2(x) - sin^2(x) ) = 2cos^2(x) - 1
( cos^2(x) + 1 - cos^2(x) )( cos^2(x) - ( 1 - cos^2(x) ) ) = 2cos^2(x) - 1
( 1 )( 2cos^2(x) - 1 ) = 2cos^2(x) - 1
~
tan(θ) + cot(θ) = 2csc(2θ)
tan(θ) + cot(θ) = 2/sin(2θ)
tan(θ) + cot(θ) = 1/sin(θ)cos(θ)
sin(θ)/cos(θ) + cos(θ)/sin(θ) = 1/sin(θ)cos(θ)
sin^2(θ)/cos(θ)sin(θ) + cos^2(θ)/sin(θ)cos(θ) = 1/sin(θ)cos(θ)
( sin^2(θ) + cos^2(θ) )/sin(θ)cos(θ) = 1/sin(θ)cos(θ)
1/sin(θ)cos(θ) = 1/sin(θ)cos(θ)

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1 - sin(t) = √3)cos(t)
1 - 2sin(t) + sin^2(t) = 3cos^2(t)
1 - 2sin(t) + sin^2(t) = 3(1 - sin^2(t))
1 - 2sin(t) + sin^2(t) = 3 - 3sin^2(t)
4sin^2(t) - 2sin(t) - 2 = 0
2sin^2(t) - sin(t) - 1 = 0
(2sin(t) + 1)(sin(t) - 1) = 0
2sin(t) + 1 = 0 and sin(t) - 1 = 0
sin(t) = -1/2 and sin(t) = 1

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sin(b) + 2cos^2(b) = 1
sin(b) + 2(1 - sin^2(b)) = 1
sin(b) + 2 - 2sin^2(b) = 1
2sin^2(b) - sin(b) - 1 = 0
* when you factor, you treat this much like any other polynomial .. for understanding sakes, x = sin(b)
2x^2 - x - 1 = 0
2x^2 - 2x + x - 1 = 0
(2x^2 - 2x) + (x - 1) = 0
2x(x - 1) + (x - 1) = 0
(2x + 1)(x - 1) = 0
* x = sin(b)
(2sin(b) + 1)(sin(b) - 1) = 0
2sin(b) + 1 = 0 and sin(b) - 1 = 0
sin(b) = -1/2 and sin(b) = 1

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Let's begin with angle O. We need a trig identity that would include the opposite leg and the hypotenuse. _SOH_ CAH TOA
sin O = x / 2.5x
sin O = 1 / 2.5
sin O = 0.4
O = 23.6 degrees
Angle Q = 90 - Angle O
Angle Q = 90 - 23.6
Angle Q = 66.4 degrees

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Judy is driving along a highway that is climbing a steady 9-degree slope. After driving for 2 miles along this road, how much altitude has Judy gained?

This has to do with a leg and the hypotenuse, so tangent is ruled out. Since this involves the opposite leg and the hypotenuse, sine is required.
sin( 9* ) = x / 2
x = 0.31 miles
How far must she travel in order to gain a mile of altitude?

sin( 9* ) = 1 / x
x = 6.39 miles

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x^x^x^x^.....x^3 = 3
log( x^x^x^x^.....x^3 ) = log( 3 )
x^x^x^.....x^3 * log( x ) = log( 3 )
3 * log( x ) = log( 3 )
log( x ) = log( 3 )/3
x = 10^( log( 3 )/3 )
About 1.4422

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e^-x = 1 / e^x
Transformation: (x,y) ~> (-x,y) so flip about y-axis
Horizontal asymptote is the value for y when x approaches infinity, or increases indefinately.
1 / e = 0.37
1 / e^2 = 0.14
1 / e^3 = 0.05
1 / e^4 = 0.02
Horizontal Asymptote: y = 0
Y-intercept as coordinates (0,y)
y = 1 / e^x
y = 1 / e^0
y = 1
So: (0,1)

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b.)
For example:
g(x) = 1 / x
g(3x) = 1 / (3x) or g(x/3) = 3 / x

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log 6
log 2*3
log 2 + log 3
0.301 + 0.477
0.778
a.)

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( 20 * 10% + x * 100% ) / ( 20 + x ) = 25%
( 20 * 0.1 + x ) / ( 20 + x ) = 0.25
( 2 + x ) / ( 20 + x ) = 0.25
2 + x = 0.25( 20 + x )
2 + x = 5 + 0.25x
0.75x = 3
x = 4kg salt

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y = log π
10^y = 10^log π
* 10^log π = π
10^y = π
c.)

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I'm guessing you are adding 100% or pure antifreeze.
( 80 * 40% + x * 100% ) / ( 80 + x ) = 52%
( 32 + x ) / ( 80 + x ) = 0.52
32 + x = 0.52( 80 + x )
32 + x = 41.6 + 0.52x
0.48x = 9.6
x = 20 Litres of antifreeze

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i^(-35)
1/i^35
1/( i * i^34 ) ~ you want an even power
1/( i * (i^2)^17 ) ~ now, you want an exponent of 2
1/( i * (-1)^17 ) ~ i^2 = -1
-1 / i ~ (-1)^17 = -1
..
4 / (2 + 5i)
4(2 - 5i) / (2 + 5i)(2 - 5i) ~ multiply the num. and dem. by the conjugate

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c

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b

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c.
because there is only one sign change for f(x)

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Plug x = -3 and y = 4 into each:
-15 + 28 = b
-3m + 4 = 22
so:
13 = b
m = -6
now:
5x + 7y = 13
-6x + y = 22

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There are two answers: solving normally, and solving when you negate
1. | x | = 6
x = 6 and x = -6
2. |3x + 2| = 14
3x + 2 = 14 and 3x + 2 = -14
3x = 12 and 3x = -16
x = 4 and x = -16/3
3. -5| x + 1 | = -10
| x + 1 | = 2
x + 1 = 2 and x + 1 = -2
x = 1 and x = -3
... and so on ...

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4x^2 – 4x + 5 = 0
4x^2 + (-4)x + 5 = 0
ax^2 + bx + c = 0
a = 4
b = -4
c = 5
(i) Compute the discriminant, b2 – 4ac, and then state whether one real-number solution, two different real-number solutions, or two different imaginary-number solutions exist.
disc. > 0 ~ two real
disc. = 0 ~ one real
disc. < 0 ~ two imaginary
(ii) Use the quadratic formula to find the exact solutions of the equation.

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The altitude makes a right triangle:
a^2 + b^2 = c^2
a^2 + 4^2 = 10^2
a^2 + 16 = 100
a^2 = 84
a = 2*sqrt(21)
Base is twice "a":
AC = 4*sqrt(21)

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A = P*(1 + r)^t
A = 45,103,674*(1.014)^4

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You first try to factor the quadratic:
2x^2 + 3x + 6 = 0
We need factors of 2 * 6 or 12 that sum to 3, which there are none.
The simplest way from here is to use the quadratic formula:

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The solution is imaginary because of sqrt( negative ) .. you can simplify from here.

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That should be right. Your line should be a perpendicular bisector to the segment formed by the two points.
m = (4 - 2) / (-1 - 5) = -1 / 3
Perpendicular slope = 3
w/ point ( (-1 + 5)/2 , (4 + 2)/2 ) or ( 2 , 3 )
y - y1 = m(x - x1)
y - 3 = 3(x - 2)
y - 3 = 3x - 6
y = 3x - 3

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AB and CD are parallel segements cut by transversal EF with G at point of intersection of AB and EF and H at CD and EF:
angle EGB = angle GHD ~ corresponding angles
* angle EGB = angle AGH ~ vertical angles
angle AGH = angle GHD
Now, a repeat of the properties would show the other two interior angles are congruent.