Answer:
Suppose we have a random number A.
The multiplicative inverse of A is a number X such that:
A*X = 1
When we work with real numbers, X = 1/A
Then:
A*(1/A) = A/A = 1
This means that (1/A) is the multiplicative inverse of A.
Where we need to have A ≠ 0, because we can not divide by 0.
Now we want to find the multiplicative inverse of the numbers:
2: Here the inverse is (1/2) = 0.5
1/5: Here the inverse is (1/(1/5)) = (5/1) = 5
-4: Herre the inverse is (1/(-4)) = -(1/4) = -0.25
0.963=0.96 rounded to the nearest hundredth
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A
rational number is any number that can be written as the
ratio between two other numbers i.e. in the form

Part A:
An easy choice that makes sense is 7.8, right in the middle. To prove that it's rational we need to write it as a ratio. In this case we have

Part B:
We need a number that can't be written as a ratio (because it neither terminates nor repeats). Some common ones are

,

,

and

so it makes sense to try and use those to build our number. In this case

works nicely.
Answer:
![9c^3 - 12c^2 - 18c - 24= [3c^2 - 6][3c - 4]](https://tex.z-dn.net/?f=9c%5E3%20-%2012c%5E2%20-%2018c%20-%2024%3D%20%5B3c%5E2%20-%206%5D%5B3c%20-%204%5D)
Step-by-step explanation:
Given

Required
Factor
Group into 2
![[9c^3 - 12c^2] - [18c + 24]](https://tex.z-dn.net/?f=%5B9c%5E3%20-%2012c%5E2%5D%20-%20%5B18c%20%2B%2024%5D)
Factorize each group
![3c^2[3c - 4] - 6[3c - 4]](https://tex.z-dn.net/?f=3c%5E2%5B3c%20-%204%5D%20-%206%5B3c%20-%204%5D)
Factor out 3c - 4
![[3c^2 - 6][3c - 4]](https://tex.z-dn.net/?f=%5B3c%5E2%20-%206%5D%5B3c%20-%204%5D)
Hence:
![9c^3 - 12c^2 - 18c - 24= [3c^2 - 6][3c - 4]](https://tex.z-dn.net/?f=9c%5E3%20-%2012c%5E2%20-%2018c%20-%2024%3D%20%5B3c%5E2%20-%206%5D%5B3c%20-%204%5D)