Answer:
Numerator = 2(b^2+a^2) or equivalently 2b^2+2a^2
Denominator = (b+a)^2*(b-a), or equivalently b^3+ab^2-a^2b0-a^3
Step-by-step explanation:
Let
S = 2b/(b+a)^2 + 2a/(b^2-a^2) factor denominator
= 2b/(b+a)^2 + 2a/((b+a)(b-a)) factor denominators
= 1/(b+a) ( 2b/(b+a) + 2a/(b-a)) find common denominator
= 1/(b+a) ((2b*(b-a) + 2a*(b+a))/((b+a)(b-a)) expand
= 1/(b+a)(2b^2-2ab+2ab+2a^2)/((b+a)(b-a)) simplify & factor
= 2/(b+a)(b^2+a^2)/((b+a)(b-a)) simplify & rearrange
= 2(b^2+a^2)/((b+a)^2(b-a))
Numerator = 2(b^2+a^2) or equivalently 2b^2+2a^2
Denominator = (b+a)^2*(b-a), or equivalently b^3+ab^2-a^2b0-a^3
Horizontal, it’s below a 1/2 slope.
In order of problems/numbers - 4, 3, 5, 6, 1, 2
Answer:

Explanation:
Here, we want to use the factor theorem to check if the given linear expression is a factor of the binomial
Now, according to the factor theorem, a factor of a polynomial would leave no remainder when divided by it
Mathematically, it means when we substitute the factor value into the polynomial, it is expected that the remainder is zero is the substituted is a factor of the polynomial
We set x-2 to zero:

Now, we substitute 2 into the polynomial as follows:

There is a remainder of -28 and thus, the linear factor is not a factor of the binomial
A)
Let x represent the cost of 1 student, and y the cost of 1 teacher.
B)
In the first group, there's 25 students and 2 teachers. Their total cost is $97.50
So 25x + 2y = 97.50
In the second group, there's 32 students and 3 teachers. Their total cost is $127
So 32x + 3y = 127
We get the following system of equations:
25x + 2y = 97.50 (1)
32x + 3y = 127 (2)
C)
25x + 2y = 97.50 (1)
32x + 3y = 127 (2)
In equation (1)
25x + 2y = 97.50
25x + 2y - 2y = 97.50 - 2y
25x = 97.50 - 2y
25x / 25 = 97.50/25 - 2y/25
x = 3.9 - (2/25)y
In equation (2), let's replace x by its algebraic value
32x + 3y = 127
32(-2/25y + 3.9) + 3y = 127
11/25y + 124.8 = 127
11/25y + 124.8 - 124.8 = 127 - 124.8
11/25y = 2.2
(11/25y) / (11/25) = 2.2 / (11/25)
y = 5
x = -2/25y + 3.9
x = -2/25 * 5 + 3.9
x = 3.5
So the cost of each student is $3.5, and the cost of each teacher is $5.
Hope this helps! :)