Given problem;
A = 4 π R²
To solve for R, we have to make it the subject of the expression.
Since
A = 4 π R² , follow these steps to find R;
Multiply both sides by 
A x = x 4 π R²
= R²
Then find the square root of both sides
√R² = 
R =
The solution is
Answer:
Step-by-step explanation:
Let, 
= y
sin(y) = 
---------(1)
cos(y) = 
= 
= 
Therefore, from equation (1),
Or ![\frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Ctext%7Bsin%7D%5E%7B-1%7D%28%5Cfrac%7Bx%7D%7B6%7D%29%5D%3D%5Cfrac%7B1%7D%7B6%5Csqrt%7B1-%5Cfrac%7Bx%5E2%7D%7B36%7D%7D%7D)
At x = 4,
![\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Ctext%7Bsin%7D%5E%7B-1%7D%28%5Cfrac%7B4%7D%7B6%7D%29%5D%3D%5Cfrac%7B1%7D%7B6%5Csqrt%7B1-%5Cfrac%7B4%5E2%7D%7B36%7D%7D%7D)
![\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Ctext%7Bsin%7D%5E%7B-1%7D%28%5Cfrac%7B2%7D%7B3%7D%29%5D%3D%5Cfrac%7B1%7D%7B6%5Csqrt%7B1-%5Cfrac%7B16%7D%7B36%7D%7D%7D)
Answer:
p = 6
Step-by-step explanation:
3p - 4 = 14 / +4
3p = 18 / ÷ 3
p = 6
Answer:
x = - 2, x = 6
Step-by-step explanation:
Given f(x) = 18 we require to solve
3 | x - 2 | + 6 = 18 ( subtract 6 from both sides )
3 | x - 2 | = 12 ( divide both sides by 3 )
| x - 2 | = 4
The absolute value function always returns a positive value, however, the expression inside can be positive or negative, thus
x - 2 = 4 ( add 2 to both sides )
x = 6
OR
- (x - 2) = 4
- x + 2 = 4 ( subtract 2 from both sides )
- x = 2 ( multiply both sides by - 1 )
x = - 2
As a check substitute these values into the left side of the equation and if equal to the right side then they are the solutions
x = 6 → 3|6 - 2| + 6 = 3|4| + 6 = 3(4) + 6 = 12 + 6 = 18 ← True
x = - 2 → 3|- 2 - 2| + 6 = 3|-4| + 6 = 3(4) + 6 = 12 + 6 = 18 ← True
Hence solutions are x = - 2, x = 6
