<span>, y+2 = (x^2/2) - 2sin(y)
so we are taking the derivative y in respect to x so we have
dy/dx use chain rule on y
so y' = 2x/2 - 2cos(y)*y'
</span><span>Now rearrange it to solve for y'
y' = 2x/2 - 2cos(y)*y'
0 = x - 2cos(y)y' - y'
- x = 2cos(y)y' - y'
-x = y'(2cos(y) - 1)
-x/(2cos(y) - 1) = y'
</span><span>we know when f(2) = 0 so thus y = 0
so when
f'(2) = -2/(2cos(0)-1)
</span><span>2/2 = 1
</span><span>f'(2) = -2/(2cos(0)-1)
cos(0) = 1
thus
f'(2) = -2/(2(1)-1)
= -2/-1
= 2
f'(2) = 2
</span>
Answer:
Part 1) 
Part 2) 
Part 3) 
Step-by-step explanation:
step 1
Find the measure of side c
Applying the law of cosines

we have

substitute





step 2
Find the measure of angle A
Applying the law of sines

substitute the given values

solve for sin(A)



step 3
Find the measure of angle B
we know that
The sum of the interior angles in any triangle must be equal to 180 degrees
so

substitute the given values



Wait can someone answer this I need help too
Answer:
y=3x-2
Step-by-step explanation:
Because the equation for slope intercept form is y=mx+b!
Answer:
For question 3, you would just add 2 to the x values and subtract 2 from the y values, so it would be:
J' (-2, 5)
K' (2, 6)
L' (1, 2)
M' (-3, 1)
For question 4 you would subtract 7 from the x values and 6 from the y values, and that would be:
W' (-6, 1)
X' (-1, -1)
Y' (-3, -6)
Z' (-8, -4)
For question 9 you would end up with:
X' (6, -5)
Y' (7, 1)
Z' (4, 0)
For question 10 you would end up with:
Q' (-1, 2)
R' (1, 7)
S' (-2, 6)
T' (-4, 1)
For question 11 you would end up with:
L' (4, 1)
M' (8, 5)
N' (6, 7)
P' (2, 3)
For question 12 you would end up with:
G' (6, -7)
H' (6, -4)
I' (1, -7)
Hope this is what you were looking for!
Step-by-step explanation: