Answer:
The answer is $88
Step-by-step explanation:
The Math.random () function returns a floating-point, pseudo-random number in the range 0–1 (inclusive of 0, but not 1) with approximately uniform distribution over that range — which you can then scale to your desired range.
Answer:
Bottom answer
Step-by-step explanation:
Although I cannot see the answer labels, the answer is the very bottom graph. The way you can find a functional graph is if you add vertical lines going down and that there is only one point that crosses it, not two. It is kind of difficult for me to explain, so if you need help, I suggest you look up how to know when a graph is a function or not.
Answer:
24 in³
Step-by-step explanation:
The volume of a prism is given by the formula ...
V = Bh
where B is the area of the base, and h is the height.
The volume of a pyramid is given by the formula ...
V = (1/3)Bh
where B and h have the same definitions.
For a pyramid with the same B and h as a prism, the volume is 1/3 that of the prism:
(1/3) × (72 in³) = 24 in³ . . . . volume of the pyramid
You can use the Pythagorean Theorem. Which is a^2 + b^2 = c^2.
So now to plug in the equation.
In your case a is 8 and c is 34. So now we need to solve for b.
8 squared is equal to 64.
34 squared is equal to 1156.
Now to plug things in.
64 + b^2 = 1156
Now subtract 64 from 1156.
That equals 1092.
The equation is now
b^2 = 1092
Now do the square root of 1092.
But since the square root is an irrational number you can just write b = 1092 ( but put a square root with 1092)
If you're using the app, try seeing this answer through your browser: brainly.com/question/2264253_______________
Evaluate the indefinite integral:
Trigonometric substitution:
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}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Blcl%7D%20%5Cmathsf%7Bx%3Dsin%5C%2C%5Ctheta%7D%26%5Cquad%5CRightarrow%5Cquad%26%5Cmathsf%7Bdx%3Dcos%5C%2C%5Ctheta%5C%2Cd%5Ctheta%5Cqquad%5Ccheckmark%7D%5C%5C%5C%5C%5C%5C%20%26%26%5Cmathsf%7Bx%5E2%3Dsin%5E2%5C%2C%5Ctheta%7D%5C%5C%5C%5C%20%26%26%5Cmathsf%7Bx%5E2%3D1-cos%5E2%5C%2C%5Ctheta%7D%5C%5C%5C%5C%20%26%26%5Cmathsf%7Bcos%5E2%5C%2C%5Ctheta%3D1-x%5E2%7D%5C%5C%5C%5C%20%26%26%5Cmathsf%7Bcos%5C%2C%5Ctheta%3D%5Csqrt%7B1-x%5E2%7D%5Cqquad%5Ccheckmark%7D%5C%5C%5C%5C%5C%5C%20%26%26%5Ctextsf%7Bbecause%20%7D%5Cmathsf%7Bcos%5C%2C%5Ctheta%7D%5Ctextsf%7B%20is%20positive%20for%20%7D%5Cmathsf%7B%5Ctheta%5Cin%20%5Cleft%5B%5Cdfrac%7B%5Cpi%7D%7B2%7D%2C%5C%2C%5Cdfrac%7B%5Cpi%7D%7B2%7D%5Cright%5D.%7D%20%5Cend%7Barray%7D)
So the integral
becomes
Integrate
by parts:
Substitute back for the variable x, and you get
I hope this helps. =)
Tags: <em>integral inverse sine function angle arcsin sine sin trigonometric trig substitution differential integral calculus</em>
