What is the first derivative of p with respect to q (i. e., differentiate P with respect to q)?P = 6q2 + 3

Step-by-step explanation:

Given P = 6q² + 3, we are to find the derivative of the function with respect to q. Generally if y = axⁿ;

dy/dx = naxⁿ⁻¹

Rewriting the given equation as P =  6q² + 3q⁰

Using the formula to find the derivative;

dP/dq = 2(6)q²⁻¹ + 0(3)q⁰⁻¹

dP/dq = 12q¹ + 0

dP/dq = 12q

Hence the derivative of the function P with respect to q is 12q

the solution is 12x-6y

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47