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andreyandreev [35.5K]
3 years ago
7

How do I find the sum of 3x8

Mathematics
2 answers:
kakasveta [241]3 years ago
5 0
You add eight three times (8+8+8=24)
tensa zangetsu [6.8K]3 years ago
3 0

Answer:

You may add eight three times or eight three times

Step-by-step explanation:

3+3+3+3+3+3+3+3=24

8+8+8=24

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Tourists from a tour bus were asked about places they visited during their stay in a city. The results are shown in the table.
Eddi Din [679]
The answer is probaly 60 % i think
3 0
4 years ago
HELP PLEASE SOMEONE HELP MEEE ASAPPPP
diamong [38]

Answer:

5

Step-by-step explanation:

You will multiply 1/3 by 3. This equals 1. 5^1 equals 5.

7 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
21 times 9 equals 9 times 21 what is the property
Reil [10]
Commutative property....
5 0
3 years ago
Please help! I really need help on this. Also if I could get an explanation on this it would really help.
Firdavs [7]

Answer:

x= 10

Step-by-step explanation:

Since the triangles are congruent

<B = <E

50 = 5x

Divide each side by 5

50/5 = 5x/5

10 =x

4 0
3 years ago
Read 2 more answers
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