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
1. 56 : 35 = 1.6
2. 1.6* 100 = 160
3. The answer is 160
Let, the first number = C
second number will be = C + 2 ['cause next will be odd so we can't include that]
Third number = C + 4
Now, their sum would be:
C + C+2 + C+4 = 3C + 6
In short, Your Answer would be 3C + 6
Hope this helps!
Answer:
5 Squash, 10 Tomatoes
Step-by-step explanation:
10 is double 5 :/
