Answer:
answer is 100 pages per hour
I am looking on the answers, and there is only one case, when a or b or c or d pass: 3|x-3| + 2 = 14. So I assume, that before two is plus. Then:
3|x-3|+2=14 |minus 2
3|x-3|=12 |divide 3
|x-3|=4
From absolute value definition you've got two ways:
x-3=4 or x-3=-4
x=7 or x=-1
And answer d) passes
I'm guessing your problem is this:
y³ - 9y² + y - 9 = 0
right?
In solving this problem, I recommend doing this:
y³ - 9y² + y - 9 = 0
Factor out a y² from the first two numbers in the problem:
y²(y - 9) + (y - 9) = 0
Separate the parentheses which means y - 9 goes on one side. The y² added a one since it came from the + 1 in the middle of expression. When you're separating parentheses like this you just take the outside numbers and combine them together. Since + 1 came from the outside of the (y - 9) and y² also was sitting on the outside of (y - 9) combine them to make y² + 1. Like this:
(y² + 1)(y - 9) = 0
Now separate your two parentheses to two separate problems:
(y² + 1) = 0 and (y - 9) = 0
Now you're y² + 1 will equal:
y² = -1
y = √-1 <-- This number doesn't exist so it will be an imaginary number (i). If you guys didn't learn that in your class I recommend just leaving it as i for that part.
Now solve y - 9 = 0:
y = 9 <-- Since we added nine to both sides to get this.
So you're final answer should be y = i and 9
i calculated this problem and it's just 46.656
F(5)= 3.(5)+2=15+2=17
Another example: F(0)=3.(0)+2=0+2=2
You just replace the value in the two sides.
