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3 years ago

What makes composing functions different from doing an arithmetic operation of two functions such as adding or dividing them?

1 answer:

3 0

function composition is an operation that takes two functions f and g and produces a function h such that h(x) = g(f(x)). In this operation, the function g is applied to the result of applying the function f to x.

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Isabella earns $48.75 for 5 hours of babysitting. At this rate, how much more would she earn for 11 hours of babysitting?

MAVERICK [17]

She would earn $107.25

4 0

3 years ago

Read 2 more answers

Find the missing variable and indicated angle measure<br> Please help lol

Margarita [4]

Answer:

Hello! x = 6

angle measure: 64

HOPE THAT HELPS a bit when trying to solve the whole thing

8 0

3 years ago

Find the LCM of the following

Tresset [83]

Answer:

A. LCM of 6,9, and 15 is 90.

B. LCM of 3,12, and 16 is 48.

C. LCM of 8,9, and 12 is 72.

D. LCM of 24 and 18 is 72.

E. LCM of 21 and 28 is 84.

F. LCM of 19 and 42 is 798.

Step-by-step explanation:

Hope this helps! :)

8 0

3 years ago

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

4 years ago

Evaluate the function for x = 2 and x = 6.<br> f(x) = -(x - 2)<br> need help like asap

Kay [80]

Answer:

zero and negative four

I hope that's what you meant I don't know

Step-by-step explanation:

f(x) = -(x - 2)

=-(2-2)

f(x) = -(x - 2)

=-(6-2)

=-4

5 0

3 years ago