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Lena [83]
3 years ago
13

What is the inverse of the function Fox(x) = 2x + 1? ​

Mathematics
1 answer:
AysviL [449]3 years ago
4 0

Answer:

f ^−1(x)=x/2−1/2

Step-by-step explanation:

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Explain why (1/2) sqaured 4 = 1/16
natka813 [3]

Step-by-step explanation:

(\frac{1}{2})^4 = (\frac{ 1^4 }{2^4}) = (\frac{1}{16})

HOPE THIS HELPS

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What’s multiplies to get 18 but adds to get -15
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Find the volume of this sphere.<br> Use 3 for TT.<br> V V ~ [?]cm3<br> V = nr3<br> r=6cm
Katen [24]

Answer:

V = 864cm^3

Step-by-step explanation:

Given

r = 6

\pi = 3

Required

The volume of the sphere

This is calculated as:

V = \frac{4}{3} \pi r^3

So, we have:

V = \frac{4}{3} *3 *6^3

V = \frac{4}{3} *3 *216

This gives:

V = 4 *216

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6 0
3 years ago
A number plus 12 is 36. write an equation and use equation to solve
just olya [345]

Answer:

EQUATION IS GIVEN BELOW.

Step-by-step explanation:

let the number be 'x'

now,

x+12 =36

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therefore,x=24

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3 years ago
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Suppose theta is an angle in the standard position whose terminal side is in quadrant 4 and cot theta = -6/7. find the exact val
zimovet [89]

First off, let's notice that the angle is in the IV Quadrant, where sine is negative and the cosine is positive, likewise the opposite and adjacent angles respectively.

Also let's bear in mind that the hypotenuse is never negative, since it's simply just a radius unit.

\bf cot(\theta )=\cfrac{\stackrel{adjacent}{6}}{\stackrel{opposite}{-7}}\qquad \impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{6^2+(-7)^2}\implies c=\sqrt{36+49}\implies c=\sqrt{85} \\\\[-0.35em] ~\dotfill

\bf tan(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{adjacent}{6}} ~\hfill csc(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{opposite}{-7}} ~\hfill sec(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{adjacent}{6}} \\\\\\ sin(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{sin(\theta)=\cfrac{-7}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies sin(\theta)=-\cfrac{7\sqrt{85}}{85}}

\bf cos(\theta)=\cfrac{\stackrel{adjacent}{6}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{cos(\theta)=\cfrac{6}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies cos(\theta)=\cfrac{6\sqrt{85}}{85}}

6 0
3 years ago
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