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:





6.2:
fraction: .2 = 0.20 = 20/100
fraction: 6 20/100
Word form: six point two
two and five hundredths
fraction: 2 5/100
decimal form 2.05
hope this helps
Ok so first you need to put w and x in the equation —-> 13-0.5(10)+6(1/2) and now you just solve -0.5 times 10 is -5 so 13-(-5)+6(1/2) now multiply 6 times 1/2 which is 3 so now we have 13-(-5)+3= 21
Cos ø = Base/ Hypotenuse
Cos 30 = y/32
Root 3/2 = y/32
y= 16root3
X= 16