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3 years ago

HELP PLZZ!!!! (pic attached)

Mathematics

1 answer:

iVinArrow [24]3 years ago

3 0

cual se sipone que hay que responder?

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GEOMETRY HELP!

stiks02 [169]

The answer is 12. I would do it by setting up a proportion:

8 = x

10 15

Then cross multiply to get

10x=8*15

10x=120

x=12

3 0

4 years ago

An ice cream machine produced 52 ice creams per minute. After reconditioning, its speed increased to 65 ice creams per minute. B

Anastasy [175]

Answer:25 %

Step-by-step explanation:

Given

Earlier machine was producing 52 ice cream per minute and

Now it is Producing 65 ice creams per minute

So percentage increase of the machine $=\frac{\text{final-Initial}}{\text{Initial}}\times 100$

$=\frac{65-52}{52}\times 100$

$=\frac{13}{52}\times 100$

$=\frac{1}{4}\times 100=25\ \%$

So there is increase of 25 % in speed

5 0

3 years ago

Wendy's hair was 45 cm long. She got 19% cut off. How long is her hair now?

Ira Lisetskai [31]

Answer:

36.45 cm

Step-by-step explanation:

You can automatically assume that if it was 20% cut off, Wendy's hair would've been 36 cm. Multiply 45 by 0.19 to get 8.55 since that is how much Wendy's hair is being cut off. Subtract that number from 45 to get your final answer which is 36.45 cm.

5 0

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. =)

6 0

3 years ago

In how many ways can 3 singers be selected from 5 who came to an audition?. A. 1. B. 10. C. 5. D. 60

pychu [463]

Answer:

Answer:

(5 3 ) = 10

Explanation:

This is a combination problem - we don't care about the order in which the singers are selected:

C n , k = ( n k ) = n !

( k ! ) ( n − k ) ! with n = population ,

k = picks

( 5 3 ) = 10

Step-by-step explanation:

4 0

3 years ago

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