What is the length of LJ? the choices are 23.0 17.0 4.7 3.5

What is 8% of 37...........

docker41 [41]

Answer:

2.96

Explanation:

37/x=100/8

(37/x)*x=(100/8)*x - we multiply both sides of the equation by x

37=12.5*x - we divide both sides of the equation by (12.5) to get x

37/12.5=x

2.96=x

x=2.96

7 0

2 years ago

Read 2 more answers

Simplify 4+(9×2)÷(4-2)+1

kogti [31]

First we have to work out everything inside the parentheses (according to PEMDAS):

9 × 2 = 18

4 - 2 = 2

Now, we can rewrite the expression:

4 + 18 ÷ 2 + 1

According to PEMDAS, we should complete the division in the middle first:

18 ÷ 2 = 9

Now we can rewrite the expression again:

4 + 9 + 1

And now it's as simple as just adding the three numbers together:

4 + 9 + 1 = 14

Your final answer is 14. :)

3 0

3 years ago

Read 2 more answers

A jogger can jog 3 miles in 15 minutes and 18 miles in 1 1/2 hours. If the number of miles and the time are in a proportional li

Anastaziya [24]

Answer: He jogs 12 miles per hour

Step-by-step explanation: Every 15 minutes he jogs 3 miles, there are 60 minutes in an hour, 60/15 = 4, 4 x 3 = 12

8 0

3 years ago

Kelly went to a store to purchase a coffee pot. She will use a coupon for 20% off. She can calculate the cost before sales tax u

galina1969 [7]

A had it before tee her

8 0

2 years ago

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

3 years ago