Rom

you're right though I'd do it as

\(x^2+a x+b - (x^2+c x+d) = 0\\ (a-c)x + (b-d) = 0 \\ \forall a \neq c \\ x = \dfrac{d-b}{a-c}\)

It's not

I assume that the interest only accrues once a year.

So we have two scenarios.  Angela applies $500 to either the credit card debt, or to the student loan debt.

First scenario the credit card is paid off.  The student loan debt increases to (1+0.07) x $2000 = $2140

Second scenario the credit card debt increases to (1+0.15) x $500 = $575

and the loan debt increases to (1+0.07) x $1500 = $1605

for a total debt of $575 + $1605 = $2180

So she is $2180 - $2140 = $40 better off paying the credit card off in full first

\(x^{-3/2} + 2x^{-1/2}+x^{1/2} =\\ x^{1/2}\left(x^{-2}+2x^{-1}+1\right) =\\ x^{1/2}\left(x^{-1}+1\right)^2\)

Any product made up of pairs of like prime factors will be a perfect square

there are 2 unique pairs, (2,2), (3,3)

there are 3 unique tetrads (2,2,2,2), (2,2,3,3), (3,3,3,3)

there are 3 unique hexads (2,2,2,2,3,3), (2,2,3,3,3,3), (3,3,3,3,3,3)

there are 2 unique octads (2,2,2,2,3,3,3,3), (2,2,3,3,3,3,3,3)

there is 1 unique decad (2,2,2,2,3,3,3,3,3,3)

So 11 perfect square factors 1, 4, 9, 16, 36, 81, 144, 324, 729, 1296, 2916, 11664

You are correct.

He has to pay it off in full within 24 months.

\(\dfrac{\$6500}{24}=\$270.83\)

\(\dfrac a b \div \dfrac c d = \dfrac a b \times \dfrac d c\)

\(-1 ~\dfrac 7 8 =-\dfrac{15}{8} \\ -\dfrac{15}{8} \div \dfrac 3 4 = -\dfrac{15}{8} \times \dfrac 4 3 = -\dfrac{60}{24} = -\dfrac{5}{2}\)

if four is a zero then (x-4) is a factor

\(\text{so we divide }x^3+3x^2 - 18x-40 \text{ by } (x-4) \\ \text{to determine what the other factors might be.} \\ \dfrac{x^3 + 3x^2 - 18 x - 40}{x-4}= x^2+7 x+10\)

\(\text{This quotient is easy enough to factor as }\\ x^2 + 7x+10 = (x+5)(x+2) \\ \text{and thus}\\ x^3+ 3x^2-18x-40 = (x-4)(x+2)(x+5) \\ \text{this is choice B}\)