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1 year ago

Help Me please . I want to know Square root of 169 and 144 and 625 . please man fast

Mathematics

2 answers:

irga5000 [103]1 year ago

6 0

Answer:

$\sqrt{169} = 13\\\\\sqrt{144} = 12\\\\\sqrt{625} = 25$

Step-by-step explanation:

$\sqrt{169} = \left(13^2 \right)^{\tfrac 12} = 13^{\tfrac 22} = 13^1 = 13\\\\\sqrt{144} = \left( 12^2\right)^{\tfrac 12} = 12^{\tfrac 22} = 12^1 = 12\\\\\sqrt{625}= \left(25^2 \right)^{\tfrac 12} = 25^{\tfrac 22} = 25^1 = 25$

neonofarm [45]1 year ago

5 0

Answer:

13,12,25

Step-by-step explanation:

Square root of 169 is 13.

Square root of 144 is 12

Square root of 625 is 25.

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2 years ago

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2 years ago

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3 years ago

What number would complete the pattern? 107 98 90 83 77_68 65

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2 years ago

Integrate sin^-1(x) dx please explain how to do it aswell ...?

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If you're using the app, try seeing this answer through your browser: brainly.com/question/2264253

_______________

Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx\qquad\quad\checkmark}$

Trigonometric substitution:

$\mathsf{\theta=sin^{-1}(x)\qquad\qquad\dfrac{\pi}{2}\le \theta\le \dfrac{\pi}{2}}$

then,

$\begin{array}{lcl} \mathsf{x=sin\,\theta}&\quad\Rightarrow\quad&\mathsf{dx=cos\,\theta\,d\theta\qquad\checkmark}\\\\\\ &&\mathsf{x^2=sin^2\,\theta}\\\\ &&\mathsf{x^2=1-cos^2\,\theta}\\\\ &&\mathsf{cos^2\,\theta=1-x^2}\\\\ &&\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\checkmark}\\\\\\ &&\textsf{because }\mathsf{cos\,\theta}\textsf{ is positive for }\mathsf{\theta\in \left[\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right].} \end{array}$

So the integral $\mathsf{(ii)}$ becomes

$\mathsf{=\displaystyle\int\! \theta\,cos\,\theta\,d\theta\qquad\quad(ii)}$

Integrate $\mathsf{(ii)}$ by parts:

$\begin{array}{lcl} \mathsf{u=\theta}&\quad\Rightarrow\quad&\mathsf{du=d\theta}\\\\ \mathsf{dv=cos\,\theta\,d\theta}&\quad\Leftarrow\quad&\mathsf{v=sin\,\theta} \end{array}\\\\\\\\ \mathsf{\displaystyle\int\!u\,dv=u\cdot v-\int\!v\,du}\\\\\\ \mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta-\int\!sin\,\theta\,d\theta}\\\\\\ \mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta-(-cos\,\theta)+C}$

$\mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta+cos\,\theta+C}$

Substitute back for the variable x, and you get

$\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx=sin^{-1}(x)\cdot x+\sqrt{1-x^2}+C}\\\\\\\\ \therefore~~\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx=x\cdot\,sin^{-1}(x)+\sqrt{1-x^2}+C\qquad\quad\checkmark}$

I hope this helps. =)

Tags: integral inverse sine function angle arcsin sine sin trigonometric trig substitution differential integral calculus

6 0

2 years ago

Help Me please . I want to know Square root of 169 and 144 and 625 . please man fast ​

Help Me please . I want to know Square root of 169 and 144 and 625 . please man fast