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monitta
3 years ago
10

What is the inverse of the function f(x) = x – 12?

Mathematics
1 answer:
SashulF [63]3 years ago
7 0
y=x-12\\
x=y+12\\
f^{-1}(x)=x+12
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Let f(x)=x^2+2x+3 . What is the average rate of change for the quadratic function from x=−2 to x = 5?
miskamm [114]
To find the average rate of change of  given function f(x) on a given interval (a,b):

Find f(b)-f(a), b-a, and then divide your result for f(b)-f(a) by your result for b-a:


f(b) - f(a)
------------
    b-a

Here your function is f(x) = x^2 - 2x + 3.  Substituting b=5 and a=-2,
f(5) = 5^2 -2(5)+3 =?                          and f(-2) = (-2)^2 - 2(-2) + 3 = ?


Calculate           f(5) - [ f(-2) ]
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Your answer to this, if done correctly, is the "average rate of change of the function f(x) = x^2+2x+3 on the interval [-2,5]."
5 0
3 years ago
Prove that sinxtanx=1/cosx - cosx
maks197457 [2]

Answer:

See below

Step-by-step explanation:

We want to prove that

\sin(x)\tan(x) = \dfrac{1}{\cos(x)} - \cos(x), \forall x \in\mathbb{R}

Taking the RHS, note

\dfrac{1}{\cos(x)} - \cos(x) = \dfrac{1}{\cos(x)} - \dfrac{\cos(x) \cos(x)}{\cos(x)} = \dfrac{1-\cos^2(x)}{\cos(x)}

Remember that

\sin^2(x) + \cos^2(x) =1 \implies 1- \cos^2(x) =\sin^2(x)

Therefore,

\dfrac{1-\cos^2(x)}{\cos(x)} = \dfrac{\sin^2(x)}{\cos(x)} = \dfrac{\sin(x)\sin(x)}{\cos(x)}

Once

\dfrac{\sin(x)}{\cos(x)} = \tan(x)

Then,

\dfrac{\sin(x)\sin(x)}{\cos(x)} = \sin(x)\tan(x)

Hence, it is proved

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3 years ago
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Answer:

Its a 50 percent chance

Step-by-step explanation:

The reason is because there are only two options, and i believe it is a independent

3 0
3 years ago
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I am Lyosha [343]

There is no answer because there cannot be one to one function inverse.

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Answer:

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but still do not understand why peoples are asking that basic

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