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3 years ago

PLEASE HELP !!!!!!!

Mathematics

2 answers:

Gelneren [198K]3 years ago

8 0

Answer:

f^(-1) (x) = (1/2)x-1/2

Plz Give Brainliest

Svetradugi [14.3K]3 years ago

5 0

Answer:

first option

Step-by-step explanation:

let y = f(x) and rearrange making x the subject, that is

y = 2x + 1 ( subtract 1 from both sides )

y - 1 = 2x ( divide both sides by 2 )

$\frac{y-1}{2}$ = x

Change y back into terms of x with x being the inverse, that is

h(x) = $\frac{x-1}{2}$ = $\frac{1}{2}$ x - $\frac{1}{2}$

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The series and the sigma notations are $\sum\limits^4_0 3(5)^n = 3 + 15 +75 + 375 +1875$ , $\sum\limits^4_0 4(8)^n = 4 + 32 + 256+ 2048 + 16384$ , $\sum\limits^4_0 2(3)^n = 2 + 6 + 18 + 54 + 162$ and $\sum\limits^4_0 5(3)^n = 5 + 15 + 45 + 135 + 405$

<h3>How to match each series with the equivalent series written in sigma notation?</h3>

To do this, we simply expand each sigma notation.

So, we have:

$\sum\limits^4_0 3(5)^n$

Next, we set n = 0 to 4.

So, we have:

3(5)^0 = 3

3(5)^1 = 15

3(5)^2 = 75

3(5)^3 = 375

3(5)^4 = 1875

So, we have:

$\sum\limits^4_0 3(5)^n = 3 + 15 +75 + 375 +1875$

$\sum\limits^4_0 4(8)^n$

Next, we set n = 0 to 4.

So, we have:

4(8)^0 = 4

4(8)^1 = 32

4(8)^2 = 256

4(8)^3 = 2048

4(8)^4 = 16384

So, we have:

$\sum\limits^4_0 4(8)^n = 4 + 32 + 256+ 2048 + 16384$

$\sum\limits^4_0 2(3)^n$

Next, we set n = 0 to 4.

So, we have:

2(3)^0 = 2

2(3)^1 = 6

2(3)^2 = 18

2(3)^3 = 54

2(3)^4 = 162

So, we have:

$\sum\limits^4_0 2(3)^n = 2 + 6 + 18 + 54 + 162$

$\sum\limits^4_0 5(3)^n$

Next, we set n = 0 to 4.

So, we have:

5(3)^0 = 5

5(3)^1 = 15

5(3)^2 = 45

5(3)^3 = 135

5(3)^4 = 405

$\sum\limits^4_0 5(3)^n = 5 + 15 + 45 + 135 + 405$

Hence, the series and the sigma notations are $\sum\limits^4_0 3(5)^n = 3 + 15 +75 + 375 +1875$ , $\sum\limits^4_0 4(8)^n = 4 + 32 + 256+ 2048 + 16384$ , $\sum\limits^4_0 2(3)^n = 2 + 6 + 18 + 54 + 162$ and $\sum\limits^4_0 5(3)^n = 5 + 15 + 45 + 135 + 405$