Answer:
C. 4
Step-by-step explanation:
first off, we know we need the value of HJ to solve for GH. Luckily for us HJ is congruent to JK, and we have the values to solve for that using pythagorean theorum.
4^2 + JK^2 = 5^2
16 + JK^2 = 25
JK^2 = 9
JK = 3
now we can solve for GH
3^2 + GH^2 = 5^2
9 + GH^2 = 25
GH^2 = 16
GH = 4
<h3>
Answer: sin(2x)</h3>
In other words, you'll type "sin" without quotes in the green box.
Also, you'll type "2" in the gray box, which also won't have quotes.
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Explanation:
We'll use the first hint. Note the pattern of sin cos +- cos sin which matches (somewhat) with the original expression.
Specifically, we'll use this identity
sin(A - B) = sin(A)cos(B) - cos(A)sin(B)
In this case, A = 3x and B = x
So,
sin(A - B) = sin(A)cos(B) - cos(A)sin(B)
sin(A)cos(B) - cos(A)sin(B) = sin(A - B)
sin(3x)cos(x) - cos(3x)sin(x) = sin(3x - x)
sin(3x)cos(x) - cos(3x)sin(x) = sin(2x)
-
that's the subtraction sign
Answer:
(2,2) : (6,8) → m = 3/2
(-2,5) : (3,-2) → m = -7/5
Explanation:
m (slope) = (y2-y1)/(x2-x1)