Answer:
dy/dx = (x^2 - 3)^sin x [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)]
Step-by-step explanation:
y = (x^2 - 3)^sinx
ln y = ln (x^2 - 3)^sinx
ln y = sin x * ln (x^2 - 3)
1/y * dy/dx = sin x * {1 / (x^2 - 3)} * 2x + ln(x^2 - 3) * cos x
1/y dy/dx = 2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)
dy/dx = [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)] * y
dy/dx = (x^2 - 3)^sin x [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)]
Answer:
2
Step-by-step explanation:
Ok, so if we know how much wire is needed for one outlet, we must figure out how many 18.25s fit into 50 ft of wire.
18.25*2=36.5
18.25*3=54.75 which is over 50 so it cannot possibly be more than 2.
2x^2 - quadratic - monomial
-2 - constant - monomial
3x - 9 - linear - binomial
-3x^2 - 6x + 9 - quadratic - trinomial
