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Arisa [49]
2 years ago
15

Write the percent as a fraction or mixed number in simplest form. 72.5

Mathematics
2 answers:
jeyben [28]2 years ago
4 0

Answer: 72 1/2 or 725/10 as mixed number

Step-by-step explanation:

Rzqust [24]2 years ago
4 0

Answer:

72.5/100

Step-by-step explanation: Write down the percent divided by 100, then multiply both top and bottom by 10 after the decimal point, after you multiplied you will get 752/10 then simplify which the simplest form is 145/2 you with then have an improper fraction and it will be 72 1/2

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One less than the product of four and a number is 11. What is that number ?
agasfer [191]
<h3>One less than the product of four and a number is 11. What is that number ?</h3>

Answer: The number is 3

Step-by-step explanation:

We will call X the number, then 4X - 1 = 11

4X = 11 + 1 =12

X = 12/4 =3

This is the number = 3

\textit{\textbf{Spymore}}​​​​​​

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}&#10;\mathsf{x^2=sin\,t}&\quad\Rightarrow\quad&\mathsf{2x\,dx=cos\,t\,dt}\\\\&#10;&&\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)}\\\\\\&#10;\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}\\\\\\&#10;\mathsf{=\displaystyle\frac{1}{2}\int\!\f dt}\\\\\\&#10;\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)}\\\\\\&#10;=\mathsf{\dfrac{1}{2}\,arcsin(x^2)+\dfrac{1}{2}\,arccos(x^2)}\\\\\\&#10;=\mathsf{\dfrac{1}{2}\cdot \left[\,arcsin(x^2)+arccos(x^2)\right]}\\\\\\&#10;=\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
GEOMETRY: <br><br> What is a vertical angle?
Zigmanuir [339]
Pairs of angles made by intersecting lines. 
7 0
3 years ago
Read 2 more answers
Anthony has a bag that contains 4 blue candies, 6 green candies, and 10 yellow candies. * Without looking, he pulls out a piece
BigorU [14]

Answer:

0.03

Step-by-step explanation:

6 0
3 years ago
Find The art length of a sector with an area of 8 square units.
frez [133]
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\bf \textit{arc's length}\\\\&#10;s=\cfrac{\theta \pi r}{180}\quad &#10;\begin{cases}&#10;r=radius\\&#10;\theta =angle~in\\&#10;\qquad degrees\\&#10;------\\&#10;\theta =\frac{180}{\pi }\\&#10;r=4&#10;\end{cases}\implies s=\cfrac{\frac{180}{\underline{\pi} }\underline{\pi} \cdot 4}{180}\implies s=\cfrac{\underline{180}\cdot 4}{\underline{180}}&#10;\\\\\\&#10;\boxed{s=4}

if you do a quick calculation on what that angle is, you'll notice that it is exactly 1 radian, and an angle of 1 radian, has an arc that is the same length as its radius.

that's pretty much what one-radian stands for, an angle, whose arc is the same length as its radius.
6 0
3 years ago
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