Answer: ∞
Step-by-step explanation: First, find the indefinite integral F (x), and then evaluate F (∞)–F(– ∞).
Answer:

Step-by-step explanation:

Or, if you mean (r+3Q)/h=t:

Answer:

Step-by-step explanation:
<u>Linear Combination Of Vectors
</u>
One vector
is a linear combination of
and
if there are two scalars
such as

In our case, all the vectors are given in
but there are only two possible components for the linear combination. This indicates that only two conditions can be used to determine both scalars, and the other condition must be satisfied once the scalars are found.
We have

We set the equation

Multiplying both scalars by the vectors

Equating each coordinate, we get



Adding the first and the third equations:


Replacing in the first equation



We must test if those values make the second equation become an identity

The second equation complies with the values of
and
, so the solution is

Bro that's a lot, i don't know what the heck you are talking about.
The restrictions for the equation is that the denominator can not be zero. So the restrictions for x would be what values make the denominator zero.
7x^2 + 6x = 0
factor
x(7x + 6) = 0
multiply any number by 0 and you get 0. So either x = 0 or 7x+6 = 0
since there's a x in the numerator x/x = 1 so this will not be a restriction. Then the only restriction is:
7x+6 = 0
7x = -6
x = -6/7