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2 years ago

1 name a pair of complementary angles

Mathematics

1 answer:

liraira [26]2 years ago

4 0

Complementary angles add to 90 degrees.
So your answer is B. Angles 1 and 2

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Please tell me how do you simplify the expression???? x^15 · y^9 over x^8 · y^3

anyanavicka [17]

Answer:

$x^7 * y^6$

Step-by-step explanation:

In the equation: $\frac{x^{15} * y^9 }{x^8 * y^3}$

x's can be simplified and y's can be simplified!

Always remember that for exponents, you subtract the powers when dividing

so for x: 15-8 = 7

for y: 9-3 = 6

so the final answer is: $x^7 * y^6$

5 0

3 years ago

Monique has 2 less than half as many pairs of shoes as Dalia . Monique has four pairs of shoes. Which equation can be used to fi

viktelen [127]

Answer:

4m +2m = D

Step-by-step explanation:

i think i havent done this in a long time but idk

ik that dalia has 6 shoes

4 0

3 years ago

PLEASE ANSWER THIS QUESTION AND EXPLAIN IF YOU CAN

jekas [21]

Factor out -1
-1(x^2+x-6)

Factor the inner trinomial
-1(x+3)(x-2)

Final answer: B

6 0

3 years ago

Read 2 more answers

Find the sum. 7x + 15x=

BaLLatris [955]

Answer:

7x + 15x= 21x

Step-by-step explanation:

7x + 15x

= (7 + 15)x

= 22x

4 0

2 years ago

Read 2 more answers

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

Lynna [10]

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