![\bf sin(x)[csc(x)-sin(x)]~~=~~cos^2(x) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ sin(x)\left[\cfrac{1}{sin(x)}-\cfrac{sin(x)}{1} \right]\implies \underline{sin(x)}\left[\cfrac{1-sin^2(x)}{\underline{sin(x)}} \right] \\\\\\ 1-sin^2(x)\implies cos^2(x)](https://tex.z-dn.net/?f=%5Cbf%20sin%28x%29%5Bcsc%28x%29-sin%28x%29%5D~~%3D~~cos%5E2%28x%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20sin%28x%29%5Cleft%5B%5Ccfrac%7B1%7D%7Bsin%28x%29%7D-%5Ccfrac%7Bsin%28x%29%7D%7B1%7D%20%5Cright%5D%5Cimplies%20%5Cunderline%7Bsin%28x%29%7D%5Cleft%5B%5Ccfrac%7B1-sin%5E2%28x%29%7D%7B%5Cunderline%7Bsin%28x%29%7D%7D%20%5Cright%5D%20%5C%5C%5C%5C%5C%5C%201-sin%5E2%28x%29%5Cimplies%20cos%5E2%28x%29)
recall again, sin²(θ) + cos²(θ) = 1.
Question 21
Let's complete the square
y = 3x^2 + 6x + 5
y-5 = 3x^2 + 6x
y - 5 = 3(x^2 + 2x)
y - 5 = 3(x^2 + 2x + 1 - 1)
y - 5 = 3(x^2+2x+1) - 3
y - 5 = 3(x+1)^2 - 3
y = 3(x+1)^2 - 3 + 5
y = 3(x+1)^2 + 2
Answer: Choice D
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Question 22
Through trial and error you should find that choice D is the answer
Basically you plug in each of the given answer choices and see which results in a true statement.
For instance, with choice A we have
y < -4(x+1)^2 - 3
-7 < -4(0+1)^2 - 3
-7 < -7
which is false, so we eliminate choice A
Choice D is the answer because
y < -4(x+1)^2 - 3
-9 < -4(-2+1)^2 - 3
-9 < -7
which is true since -9 is to the left of -7 on the number line.
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Question 25
Answer: Choice B
Explanation:
The quantity (x-4)^2 is always positive regardless of what you pick for x. This is because we are squaring the (x-4). Squaring a negative leads to a positive. Eg: (-4)^2 = 16
Adding on a positive to (x-4)^2 makes the result even more positive. Therefore (x-4)^2 + 1 > 0 is true for any real number x.
Visually this means all solutions of y > (x-4)^2 + 1 reside in quadrants 1 and 2, which are above the x axis.
Answer:
m=1
y= 5
Step-by-step explanation:
since y=mx+b
y=1x+5
We are given to lines XY and VW. Now we need to determine the expression that correctly states that these lines are congruent. One possibility to prove that they're congruent is if they are two separate lines and:
XA is congruent to VB,
AY is congruent to BW
XA + AY = XY
VB + BW = VW
Then we can conclude that if the statements above are true, XY and VW must be congruent to each other.
Another possibility is that they are two sides of an isosceles rectangle XYVW and are opposite sides of the rectangle. <span />
Answer:
scalene
Step-by-step explanation:
When all sides aren't equal.
