Step-by-step explanation:
60 × 60= 3600
I hope this is correct
Answer:
a) Micheal's present age = 8 years old
b) The sum of their ages in 6 years time = 63 years.
Step-by-step explanation:
From the above question, we know that
Rui feng is 3 years old
We are told that
• Michael is 5 years old than Rui feng
Micheal's present age is calculated as:
= 5 + Rui feng's age
= 5 + 3 = 8 years
• Vishal is thrice as old as Micheal
Vishal's present age is calculated as
= 3(Micheal age)
= 3(8)
= 24 years.
The sum of their ages in six years time
= ( 3 + 6) +( 8 + 6) + ( 24 + 6)
= 18 + 15 + 30
= 63 years
Therefore, a) Micheal's present age = 8 years old
b) The sum of their ages in 6 years time = 63 years.
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>
The answer is 11:5 ……………….
Answer:
-4 =x
Step-by-step explanation:
ƒ(x) = 9x + 54
Let f(x) = 18
18 = 9x+54
Subtract 54 from each side
18-54 = 9x + 54-54
-36 = 9x
Divide each side by 9
-36/9 = 9x/9
-4 =x