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3 years ago

What is the multiplicative inverse of −3/5?

Mathematics

1 answer:

bonufazy [111]3 years ago

4 0

Answer:

The multiplicative inverse is -5/3

Step-by-step explanation:

Multiplicative inverse means we want to end up with 1

-3/5 * what =1

Multiply by 5 to clear the fraction

-3/5 * what *5 = 1*5

-3 * what = 5

Divide by -3 to isolate what

-3*what /-3 = 5/-3

what = -5/3

The multiplicative inverse is -5/3

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Find the critical numbers of the function. (Enter your answers as a comma-separated list. Use n to denote any arbitrary integer

lyudmila [28]

$f(\theta)=10\cos\theta+5\sin^2\theta$

then the derivative is

$f'(\theta)=-10\sin\theta+10\sin\theta\cos\theta$

Critical points occur where $f'(\theta)=0$ . This happens for

$-10\sin\theta+10\sin\theta\cos\theta=0$

$-10\sin\theta(1-\cos\theta)=0$

$\implies-10\sin\theta=0\text{ or }1-\cos\theta=0$

In the first case, we find

$-10\sin\theta=0\implies\sin\theta=0\implies\theta=n\pi$

In the second,

$1-\cos\theta=0\implies\cos\theta=1\implies\theta=2n\pi$

So all the critical points occur at multiples of $\pi$ , or $n\pi$ . (This includes all the even multiples of $\pi$ .)

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Yakvenalex [24]

Answer:

$\boxed{\dfrac{f(a+h)-f(a)}{h}=-3a^2-3ah-h^2}$

Step-by-step explanation:

$f(x)=-x^3+2\\\\\dfrac{f(a+h)-f(a)}{h}\\\\f(a+h)\to\text{substitute x = a + h}\\f(a)\to\text{substitute x = a}\\\\\dfrac{f(a+h)-f(a)}{h}=\dfrac{-(a+h)^3+2-(-a^3+2)}{h}\\\\\text{use}\ (x+y)^3=x^3+3x^2y+3xy^2+y^3\\\\=\dfrac{-(a^3+3a^2h+3ah^2+h^3)+2-(-a^3)-2}{h}\\\\=\dfrac{-a^3-3a^2h-3ah^2-h^3+2+a^3-2}{h}\\\\\text{combine like terms}$

$=\dfrac{(-a^3+a^3)-3a^2h-3ah^2-h^3+(2-2)}{h}\\\\=\dfrac{-3a^2h-3ah^2-h^3}{h}\\\\=\dfrac{h(-3a^2-3ah-h^2)}{h}\\\\=-3a^2-3ah-h^2$

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