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Stells [14]
4 years ago
12

Which of the following is the function T(a) if T^-1(a)=11a-2

Mathematics
1 answer:
SpyIntel [72]4 years ago
5 0

Answer:

T(a) =(a+2)/11

Step-by-step explanation:

The inverse of the inverse is the function

To find the inverse let T^-1(a) be y

y = 11a -2

Exchange y and a

a = 11y -2

Solve for y

Add 2 to each side

a+2 = 11y -2 +2

a+2 =11y

Divide each side by 11

(a+2)/11 =y

T(a) =(a+2)/11

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Step-by-step explanation:

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GUYS please
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Answer:

2.64

Step-by-step explanation:

0.9(x + 1.4) - 2.3 + 0.1x = 1.6

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3 years ago
Evaluate Dx / ^ 9-8x - x2^
Solnce55 [7]
It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for \sqrt x.

In that case, you want to find the antiderivative,

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}

Complete the square in the denominator:

9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2

Now substitute x+4=5\sin y, so that \mathrm dx=5\cos y\,\mathrm dy. Then

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy

which simplifies to

\displaystyle\int\frac{5\cos 
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Now, recall that \sqrt{x^2}=|x|. But we want the substitution we made to be reversible, so that

x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)

which implies that -\dfrac\pi2\le y\le\dfrac\pi2. (This is the range of the inverse sine function.)

Under these conditions, we have \cos y\ge0, which lets us reduce \sqrt{\cos^2y}=|\cos y|=\cos y. Finally,

\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C

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\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C
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3 years ago
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Step-by-step explanation:

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Step 2: Subtract 20 from both sides.

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