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frez [133]
3 years ago
14

Describe hoy to use a tens fact to find the difference for 15-8

Mathematics
1 answer:
wolverine [178]3 years ago
6 0

Right? So if you had a ten block that would represent that for a 10 so 15 minus 10 know 5 left so we would use 5 1 blocks

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Juanita has a bag of marbles. Without looking, she removes one marble, notes the color and replaces it. She repeats this process
Svetllana [295]

Given table:

color ----- Frequency

Red ------ 14

Green ----- 19

Blue -------- 21

Yellow ----- 16

Number of process repeated (trials) = 70 times

Probability (event) = number of times the event occurs / total number of trials

From the table we can see that the number of times she pick the blue marble is 21

Probability of picking a blue marble on next try = \frac{21}{70} = \frac{3}{10}

6 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
Shawn is buying a new Jet Ski
marishachu [46]

Answer:

Step-by-step explanation:

3 0
3 years ago
Pls help for brainliest
Minchanka [31]

。☆✼★ ━━━━━━━━━━━━━━  ☾

πr^2 = area

Half the diameter to get the radius

Substitute the values in:

area = 3.142 x (7)^2

area = 153.958 cm^2

Have A Nice Day ❤    

Stay Brainly! ヅ    

- Ally ✧    

。☆✼★ ━━━━━━━━━━━━━━  ☾

8 0
3 years ago
Read 2 more answers
Solve. u+6 = 18<br> a. 3<br> b. 12<br> c. 24<br> d. 108
Gekata [30.6K]
u+6=18\\&#10;u=12
4 0
3 years ago
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