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4 years ago

Can someone help me plz

Mathematics

1 answer:

zloy xaker [14]4 years ago

4 0

Answer:

isnt it 50 and 25 because ur going x2

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If (sin(()−1))p =2cos(()−1)q , show that p^2 =4q^2 (1− q^2) ?.

Papessa [141]

Likely the equation is

$\sin^{-1}p=2\cos^{-1}q$

For either side to be defined, we require $|p|\le1$ and $|q|\le1$ . Under these conditions, we can take

$\sin(\sin^{1}p)=\sin(2\cos^{-1}q)$
$\implies p=2\sin(\cos^{-1}q)\cos(\cos^{-1}q)$
$\implies p=2q\sin(\cos^{-1}q)$
$\implies p=2q\sqrt{1-q^2}$
$\implies p^2=4q^2(1-q^2)$

as required.

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4 years ago

The function h is defined by h(y)=2 y-7 . what is the value of h(8)

gtnhenbr [62]

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Given the following exponential function, identify whether the change represents

Pavel [41]

Answer:

Step-by-step explanation:

This equation represents exponential decay. Whenever the base is less than 1, the function represents decay. When the base is greater than 1, the function represents growth. In this case, the base is .97 which is less than 1, representing decay.

The formula for exponential decay is y=a(1-r)x.

r is the decay rate, expressed as a decimal.

In this case, r = .03 which represents 3%!

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2 years ago

Limit question: lim x-->pi ((e^sinx)-1)/(x-pi)

aleksandr82 [10.1K]

$\displaystyle\lim_{x\to\pi}\dfrac{e^{\sin x}-1}{x-\pi}$

Notice that if $f(x)=e^{\sin x}$ , then $f(\pi)=e^{\sin\pi}=e^0=1$ . Recall the definition of the derivative of a function $f(x)$ at a point $x=c$ :

$f'(c):=\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$

So the value of this limit is exactly the value of the derivative of $f(x)=e^{\sin x}$ at $x=\pi$ .

You have

$f'(x)=\cos x\,e^{\sin x}\implies f'(\pi)=\cos\pi\,e^{\sin\pi}=-1$

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4 years ago