Y cos x = x^2 + y^2
Taking derivative:
<span>
y' cos x - y sin x = 2x + 2y y' </span>
Simplifying:
<span>y' = (- 2x - y sin x)/(2y - cos x)</span>
Two complementary angles = 90°
X = 18+ 2y
X + y = 90
Plug first equation into second
18 + 2y +y = 90
18 + 3y = 90
18 – 18 +3y = 90 – 18
3 y = 72
3y/3 = 72/3
Y = 24
Plug y into 2nd formula
X + 24 = 90
X + 24 -24 = 90 – 24
X = 66
Answer:
C y cause it was already there plus ik super late but hey sum points
Step-by-step explanation:
First of all, observe that all fractions exist as long as b is not zero.
Multiply both numerator and denominator of the first fraction by 3:

Now all fractions have the same denominator. Under the assumption that b is not zero, multiply both sides by 3b:

Subtract 3b from both sides:

Subtract 2 from both sides:

Divide both sides by 2:
