Answer:
omg girl have you lost your mind 0please don't do something like that it's your life make it happy you are the builder of your own life
Answer:
Step-by-step explanation:
Owing represent a negative balance.
<u>So the inequality for this case is:</u>
Answer:
helloo…
Step-by-step explanation:
I probably solved it right…
(If you don't understand my writing, I will rewrite it)
Answer:
30 degrees
Step-by-step explanation:
Answer:
See explanation
Step-by-step explanation:
Prove that

1. When
we have
- in left part
- in right part

2. Assume that for all
following equality is true

3. Prove that for
the following equality is true too.

Consider left part:

Consider right part:

We get the same left and right parts, so the equality is true for
By mathematical induction, this equality is true for all n.