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3 years ago

8

You mow the lawn to earn your allowance. Each time you mow the lawn you earn $12. Write an equation for the number of dollars, d

, you earn when you mow the lawn m times.

2 answers:

ZanzabumX [31]3 years ago

6 0

<span>d=12m
Hope this answers your question!!!:)</span>

lidiya [134]3 years ago

5 0

It would simply be d(m)

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I WILL GIVE A FREAKING BRAINLIEST!!!!!!!!!!!

UkoKoshka [18]

Answer:

0.5 < t < 2

Step-by-step explanation:

The function reaches its maximum height at ...

t = -b/(2a) = -16/(2(-16)) = 1/2 . . . . . . where a=-16, b=16, c=32 are the coefficients of f(t)

The function can be factored to find the zeros.

f(t) = -16(t^2 -1 -2) = -16(t -2)(t +1)

The factors are zero for ...

x = -1 and x = +2

The ball is falling from its maximum height during the period (0.5, 2), so that is a reasonable domain if you're only interested in the period when the ball is falling.

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3 years ago

How many units up was the graph of f(x) =∛x shifted to form the translation?

Alexus [3.1K]

The <span>graph of f(x) =∛x was shifted 2 units up to form the translation.</span>

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3 years ago

Read 2 more answers

I need help fast thanks

Dmitry [639]

-18x-12=4
Iam not sure if that’s what you want but basically I multiplied the numbers inside the bracket by -6

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3 years ago

Find \(\int \dfrac{x}{\sqrt{1-x^4}}\) Please, help

ki77a [65]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2867785

_______________

Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-x^4}}\,dx}\\\\\\ \mathsf{=\displaystyle\int\! \frac{1}{2}\cdot 2\cdot \frac{1}{\sqrt{1-(x^2)^2}}\,dx}\\\\\\ \mathsf{=\displaystyle \frac{1}{2}\int\! \frac{1}{\sqrt{1-(x^2)^2}}\cdot 2x\,dx\qquad\quad(i)}$

Make a trigonometric substitution:

$\begin{array}{lcl}
\mathsf{x^2=sin\,t}&\quad\Rightarrow\quad&\mathsf{2x\,dx=cos\,t\,dt}\\\\
&&\mathsf{t=arcsin(x^2)\,,\qquad 0\ \textless \ x\ \textless \ \frac{\pi}{2}}\end{array}$

so the integral (i) becomes

$\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{1-sin^2\,t}}\cdot cos\,t\,dt\qquad\quad (but~1-sin^2\,t=cos^2\,t)}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{cos^2\,t}}\cdot cos\,t\,dt}$

$\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{cos\,t}\cdot cos\,t\,dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\f dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\,t+C}$

Now, substitute back for t = arcsin(x²), and you finally get the result:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=\frac{1}{2}\,arcsin(x^2)+C}$ ✔

________

You could also make

x² = cos t

and you would get this expression for the integral:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=-\,\frac{1}{2}\,arccos(x^2)+C_2}$ ✔

which is fine, because those two functions have the same derivative, as the difference between them is a constant:

$\mathsf{\dfrac{1}{2}\,arcsin(x^2)-\left(-\dfrac{1}{2}\,arccos(x^2)\right)}\\\\\\
=\mathsf{\dfrac{1}{2}\,arcsin(x^2)+\dfrac{1}{2}\,arccos(x^2)}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \left[\,arcsin(x^2)+arccos(x^2)\right]}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \dfrac{\pi}{2}}$

$\mathsf{=\dfrac{\pi}{4}}$ ✔

and that constant does not interfer in the differentiation process, because the derivative of a constant is zero.

I hope this helps. =)

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3 years ago

Find the volume of the prism.

Rzqust [24]

Answer:

11,968

Step-by-step explanation:

I'm not sure, but I think its 11,968

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3 years ago