Answer:
See below
Step-by-step explanation:
I will assume the given denominator is 3a as there is nothing attached to the statement.
Lets first look at the domain of 2a+b. As it is a polynomial with unknowns a and b, we know it has its domain in all real numbers for both a and b as every polynomial does. For verifying it replace a and b by any real number you can think of.
Domain = a, b belonging to R
Now, if we divide 2a+b by 3a we will have:
(2a+b)/3a
As know we don not have a polynomial we cant state that the domain will still be all the real numbers for sure. We need to go further.
We have a fraction and as every fraction the denominator CAN'T be equal to 0, so 3a MUST be different to 0. This means that a MUST be different to 0. So, now our domain changes:
domain' = a, b belonging to R and a different to 0.
<span> When raising an number to zero power you will always get 1. you raise both the numerator and the denominator</span>
Answer:
9
Step-by-step explanation:
let a = 2
b = 5
2a + b
Substituting the values,
=> 2(2) + 5
=> 4 + 5
=> 9
Step-by-step explanation:
Work done = Force × displacement
= 350 × 40
=14000J
