2. $1.75
3. $1.65
4. $1.05
5. $1.55
6. $1.20
7. $1.50
8. $1.29
Answer:


Step-by-step explanation:
<u>Equation Solving</u>
We are given the equation:
![\displaystyle x=\sqrt[3]{\frac{3y+16}{2y+9}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3y%2B16%7D%7B2y%2B9%7D%7D)
i)
To make y as a subject, we need to isolate y, that is, leaving it alone in the left side of the equation, and an expression with no y's to the right side.
We have to make it in steps like follows.
Cube both sides:
![\displaystyle x^3=\left(\sqrt[3]{\frac{3y+16}{2y+9}}\right)^3](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%5E3%3D%5Cleft%28%5Csqrt%5B3%5D%7B%5Cfrac%7B3y%2B16%7D%7B2y%2B9%7D%7D%5Cright%29%5E3)
Simplify the radical with the cube:

Multiply by 2y+9

Simplify:

Operate the parentheses:


Subtract 3y and
:

Factor y out of the left side:

Divide by
:

ii) To find y when x=2, substitute:





The answer is 5 because 16+4=20 and 3s + s = 4s so you’re left with 4s = 20 and I know that 4 times 5 is 20.
ANSWER: S=5
The subtraction theorem states that for all real numbers,

and

,

.
(To subtract, we can add the inverse.)
Thus, we can have the these two equivalent expressions.