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3 years ago

If f(x) = 10 + 2x and g(x) = 6x + 5, then (f+ g)(x) =

Mathematics

2 answers:

Alexxandr [17]3 years ago

7 0

Answer:

8x +15

Step-by-step explanation:

f(x) = 10 + 2x and g(x) = 6x + 5

(f+ g)(x) = 10 + 2x + 6x + 5

Combine like terms

= 8x +15

never [62]3 years ago

4 0

Answer:

8x+15

Step-by-step explanation:

(f+g)(x) = f(x)+g(x)

= (10+2x) + (6x+5)

= 8x+15

hope it helped, please mark me brainliest.

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For the following telescoping series, find a formula for the nth term of the sequence of partial sums

gtnhenbr [62]

I'm guessing the sum is supposed to be

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}$

Split the summand into partial fractions:

$\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}$

$1=a(5k+4)+b(5k-1)$

If $k=-\frac45$ , then

$1=b(-4-1)\implies b=-\frac15$

If $k=\frac15$ , then

$1=a(1+4)\implies a=\frac15$

This means

$\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}$

Consider the $n$ th partial sum of the series:

$S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)$

The sum telescopes so that

$S_n=\dfrac2{14}-\dfrac2{5n+4}$

and as $n\to\infty$ , the second term vanishes and leaves us with

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}=\lim_{n\to\infty}S_n=\frac17$

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3 years ago

How do I write this fraction as a decimal 14 3/5

Softa [21]

we have

$14 \frac{3}{5}$

This is a mixed number

we know that

A mixed number is a combination of a whole number and a fraction.

$14 \frac{3}{5}$ is equal to

$14 +\frac{3}{5}$

where

$14$ is the whole number

$\frac{3}{5}$ is the fraction

$\frac{3}{5}=0.60$

$14 +\frac{3}{5} =14+0.60\\ \\ =14.60$

therefore

the answer is

$14.60$

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Novay_Z [31]

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Step-by-step explanation:

Rectangular prism

V = lwh l = 8 W = 3 h = 9

= 8(3)(9)

= 216

Cube

$V = s^{3}$ where s = length of side

216 = $s^{3}$

$s = \sqrt[3]{216}$ = 6

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