Line that never touch each other
Step-by-step explanation:
You put them together to form a bigger number then break it down. sorry if it doesn't make sense.
Btw do you form the number if so I'll help just give the number and I'll break it down for you. :)
The second term of the expansion is
.
Solution:
Given expression:

To find the second term of the expansion.

Using Binomial theorem,

Here, a = a and b = –b

Substitute i = 0, we get

Substitute i = 1, we get

Substitute i = 2, we get

Substitute i = 3, we get

Substitute i = 4, we get

Therefore,



Hence the second term of the expansion is
.
Answer:
1.19
Step-by-step explanation:
You're trying to figure out what x is here.
So you'll want to isolate the x. That means, get it alone.
Since both sides are equivalent, when you subtract 2.45 from the x side, you also subtract it from the other side too.
x = 3.64 - 2.45
x = 1.19