Approaching algebraicly(Take 2 as x)
(-x)^3-x^3
Simplify
Result is -2x^3
Put back 2
-2(2)^3 = -2(8)
= -16
Let the number<span> be H T O. Ones digit = O Given that. O =10−5 ⇒ O =5. Also is given that tens digit T is 2 </span>more than<span> ones digit O ⇒ tens digit T = O +2=5+2=7. ∴ The </span>number<span> is. H 75. Given also is that "</span>number<span> is </span>less<span> than </span>200<span> and </span>greater than 100<span> " ⇒ H can take value only =1. We get our </span>number<span>as 175.</span>
In an arithmetic sequence, the difference between consecutive terms is constant. In formulas, there exists a number
such that

In an geometric sequence, the ratio between consecutive terms is constant. In formulas, there exists a number
such that

So, there exists infinite sequences that are not arithmetic nor geometric. Simply choose a sequence where neither the difference nor the ratio between consecutive terms is constant.
For example, any sequence starting with

Won't be arithmetic nor geometric. It's not arithmetic (no matter how you continue it, indefinitely), because the difference between the first two numbers is 14, and between the second and the third is -18, and thus it's not constant. It's not geometric either, because the ratio between the first two numbers is 15, and between the second and the third is -1/5, and thus it's not constant.
Note the order of the letters:
A C E G
M N P R
You read it like this:
A ≅ M
C ≅ N
CE ≅ NP
AG ≅ MR
... and so on
In the following choices:
CE ≅ NP is the true statement
Answer:

Step-by-step explanation:
Let 
So 

Let 
We are going to find 
So we are evaluating 
First step find f(-2)
Second step find f'(-2)
Third step plug in those values and apply PEMDAS!


So
