All categories

2 years ago

Given f(x)=1/x+5 and g(x)=x-2

Mathematics

1 answer:

vladimir1956 [14]2 years ago

3 0

For this case we have the following functions:

$f (x) = \frac {1} {x + 5}\\g (x) = x-2$

We must find f (g (x)). So:

$f (g (x)) = \frac {1} {(x-2) +5} = \frac {1} {x-2 + 5} = \frac {1} {x + 3}$

Finally we have to:

$f (g (x)) = \frac {1} {x + 3}$

For the function to be defined the denominator must be different from 0. That is x different from -3.

Answer:

Option B

You might be interested in

What is the volume of a sphere with a diameter of 5 in? Round your answer to the nearest hundredth of a cubic inch.

Anettt [7]

if the sphere has a diameter of 5, then its radius is half that, or 2.5.

$\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=2.5 \end{cases}\implies V=\cfrac{4\pi (2.5)^3}{3}\implies V=\cfrac{62.5\pi }{3} \\\\\\ V\approx 65.44984694978736\implies V=\stackrel{\textit{rounded up}}{65.45}$

3 0

2 years ago

You are baking brownies for your class. There are 28 total students in your class and you have baked 23 brownies. Right and solv

slavikrds [6]

Answer:

$x = 33$

Step-by-step explanation:

Given

$Students = 28$

$Brownies = 23$

$Brownies = 2\ per\ student$

Required

Determine the number of brownies left

If:

$Brownies = 2\ per\ student$

and there are 28 students

Then:

$Total\ Brownies = 2 * 28$

$Total\ Brownies = 56$

To determine the number of brownies left, we have:

$Total\ Brownies = Brownies\ Made + Brownies\ Left$

$56 = 23 + x$

Reorder

$x + 23 = 56$

Collect Like Terms

$x = 56 - 23$

$x = 33$

<em>Hence, there are 33 brownies left</em>

6 0

2 years ago

PLEASE HELP ME!!! WHOEVER HELPS ME AND GETS IT RIGHT WILL GET BRAINIEST!!!

Kamila [148]

(0,6) is the correct answer

3 0

2 years ago

Read 2 more answers

4x-10=30<br> 2n-7=30 <br> S/3+2=4

emmasim [6.3K]

Answer:

x=10 n=18.5 s=6

Step-by-step explanation:

4x-10=30

Add 10 to both sides

4x-10+10=30+10

4x=40

Divide by 4 on both sides

(4x)/4=40/4

x=10

2n-7=30

Add 7 to both sides

2n-7+7=30+7

2n=37

Divide by 2 on both sides

(2n)/2=37/2

n=18.5

(s/3)+2=4

Subtract 2 from both sides

(s/3)+2-2=4-2

(s/3)=2

Multiply by 3 on both sides

3s/3=2*3

s=6

6 0

3 years ago

Use the Taylor series you just found for sinc(x) to find the Taylor series for f(x) = (integral from 0 to x) of sinc(t)dt based

Marina CMI [18]

In this question (brainly.com/question/12792658) I derived the Taylor series for $\mathrm{sinc}\,x$ about $x=0$ :

$\mathrm{sinc}\,x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}$

Then the Taylor series for

$f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt$

is obtained by integrating the series above:

$f(x)=\displaystyle\int\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx=C+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

We have $f(0)=0$ , so $C=0$ and so

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

which converges by the ratio test if the following limit is less than 1:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}$

Like in the linked problem, the limit is 0 so the series for $f(x)$ converges everywhere.

7 0

2 years ago