Answer: -7920
Step-by-step explanation: 33x-6x40= -7920
The zeroes of <span>f(x) = x^5-12x^2+32x can be found by factoring,
</span><span>f(x) = x^5-12x^2+32x=(x-8)(x-4)
By the zero product theorem, (x-8)=0 or (x-4)=0 which means
x=8 or x=4.
So the zeroes of f(x) are S={4,8}</span>
∫ e^(3x)*(cosh(2x)dx
= ∫ [e^(3x)*(e^(2x)+e^(-2x))/2]dx
= ∫ [(e^(5x)+e^x)/2]dx
=e^(5x)/10+e^x/2+C
=(1/10)(e^(5x)+5e^x)+C
Answer:
yeah sure
Step-by-step explanation:
because I'm friendly
