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3 years ago

15

An earthworm tunnels 2 centimeters into the ground each day the earthworm tunnels at the same pace everyday how many days will i

t tack the earthworm to tunnel 14 centimeters

2 answers:

Goshia [24]3 years ago

7 0

It will take him 7 days

marusya05 [52]3 years ago

6 0

$14 \div 2 = 7 \: days$

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Soloha48 [4]

Answer:

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3 years ago

Given f(x)=x3 and g(x)= 1-5x2, fine (fog)(x) and it’s Domain

andre [41]

Answer:

Option B. $f(g(x)) = (1-5x ^ 2) ^ 3$ all real numbers

Step-by-step explanation:

We have

$f(x) = x ^ 3$ and $g(x) = 1-5x ^ 2$

They ask us to find

(fog)(x) and it's Domain

To solve this problem we must introduce the function g(x) within the function f(x)

That is, we must do f(g(x)).

So, we have:

$f(x) = x ^ 3$

$g(x) = 1-5x ^ 2$

Then:

$f(g(x)) = (1-5x ^ 2) ^ 3$

The domain of the function f(g(x)) is the range of the function $g(x) = 1-5x ^ 2$ .

Since the domain and range of g(x) are all real numbers then the domain of f(g(x)) are all real numbers

Therefore the correct answer is the option b: $f(g(x)) = (1-5x ^ 2) ^ 3$

And his domain is all real.

5 0

3 years ago

Read 2 more answers

What is the common ratio for the geometric sequence 32,8,2,1/2?<br>

abruzzese [7]

The answer you will be dividing by is 4. hope this helps

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3 years ago

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

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3 years ago

A used motorcycle is priced at $3,438. If you borrow the money for the motorcycle, your payments will be $165 per month for 24 m

bonufazy [111]

You will save $522 by paying cash. 165*24=3960 3960-3438=522

:)

6 0

3 years ago