Answer:B? l hope is B l hope helps
Step-by-step explanation:
if not help then l will answer other problem from you l hope is right
Answer:
451
Step-by-step explanation:
Okay the answer is 69 because 81-2 is 69
<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>
Because the polynomial has degree 2, we can assume that there are 2 solutions (roots), whether real or imaginary.
You can subtract 60 in order to put this in standard form
48x^2+44x-60 = 0
From there, just put a,b, and c into the quadratic formula and you're good to solve for your answers.
(-b+-sqrt(b^2-4ac))/2a
(-44+-sqrt(44^2-4(48)(-60)))/2(48)
Then solve.
There is probably a better way, but this should give you the two roots/solutions.
