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umka2103 [35]
3 years ago
15

I really need help with my take home test. The problem is f(x)=(x-7)/2, what is the value of (f°f^-1)(3). Really struggled with

this one. Options are:
1) 2x+7

2)-2

3) 3

4) x
Mathematics
2 answers:
-Dominant- [34]3 years ago
8 0
3) 3 is the answer to this equation
kiruha [24]3 years ago
5 0
Answer: 3) 3
First,We can find the inverse function of f(x) which is f^-1(x);
Let f(x)=y , x=f^-1(y)
(x-7)/2=y
(x-7)=2y
x=2y+7
Subtitute this equation x=2y+7 in the equation f^-1(y) and we have;
f^-1(y)=2y+7
change y into x,
f^-1(x)=2x+7

(f°f^-1)(3)=(f×(f^-1)(3))
=f(2×(3)+7)
=f(6+7)
=f(13)
=(13-7)/2
=6/2
=3
Hope this helps and sorry if my answer is wrong.
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viva [34]

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y=ab^x

now, we can find a and b

Initially , price is 500

so,

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500=ab^0

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now, we can plug back

y=500(b)^x

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we can use it and find b

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3 years ago
Solving Rational equations. LCD method. Show work. Image attached.
posledela

(k-2)(k-6)=k^2-2k-6k+12=k^2-8k+12

So in order to get all the fractions to have a common denominator, we need to multiply \dfrac k{k-2} by \dfrac{k-6}{k-6}, and \dfrac1{k-6} by \dfrac{k-2}{k-2}:

\dfrac k{k-2}\cdot\dfrac{k-6}{k-6}=\dfrac{k(k-6)}{(k-2)(k-6)}=\dfrac{k^2-6k}{k^2-8k+12}

\dfrac1{k-6}\cdot\dfrac{k-2}{k-2}=\dfrac{k-2}{(k-2)(k-6)}=\dfrac{k-2}{k^2-8k+12}

Now,

\dfrac4{k^2-8k+12}=\dfrac{(k^2-6k)+(k-2)}{k^2-8k+12}

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We can factorize the right side:

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