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 23 I took the test
5 divided by one half, would be 2.5, since its half of 5.
Answer:
(see image)
bottom right image
Explanation:
First try the origin (0,0) to rule out two of the graphs.
3y ≥ x - 9 3(0) ≥ (0) - 9
3 ≥ - 9
yes 3x + y > - 3 3(0) + (0) > - 3
3 > - 3
yes so the origin should be in the shaded area of the graph, which rules out the top right and bottom left graphs.
Now try a coordinate that is in the shaded area of one of the remaining graphs, but not in the other one. If it works, the graph is the one that has that point in the shaded region, and vice versa.
Try point (4, 2)
3y ≥ x - 9
3(2) ≥ (4) - 9
6 ≥ - 5
yes3x + y > - 3
3(4) + (2) > - 3
12 + 2 > - 3
14 > - 3
yesSo the graph is the bottom right one since (4, 2) is included in that shaded region.
