Could somebody help me with this math problem

what is this expression equal to? https://assessment.fcps.edu/Uploads/FCPS…


find the quotient in this expression
https://assessment.fcps.edu/Uploads/FCPS…

what is the solution to the equation?
https://assessment.fcps.edu/Uploads/FCPS…


if f(x) = x²-3x+10, what is f(-4)

thanks alot if u guys could help me with these problems/

Q1.

3x^4 y +6x^3 y -12x^7 y^5 (3x^3 y)*(x +2 - 4x^4 y^4)
____________________ = ___________________
3x^2 y 3 x^2 y

= x(x +2 -4x^4 y^4)
= -4(x^5)(y^4) + x^2 + 2x

Q.2
(1/6)(12x+30) = 5x +7 or
2x +5 = 5x + 7 or
3x = -2 or x = (-2/3)

Q.3 f(-4) = (-4)^2 -3*(-4) +10 = 16+12 +10 = 38

1) ((3x^4)+(6x^3y)-(12x^7y^5))/(3x^2y)
Factor out a 3x^2y from each term. This will leave you with

x^2+2x-4x^5y^4

2) (x^2-3x-10)/(x+2)
Factoring the initial term, we get (x-5)(x+2)/(x+2)
(x+2) cancels out leaving you with (x-5)

3) (1/6)(12x+30)=(5x+7)
Multiply the (1/6) through the terms on the left of the equals sign, and you get
2x+5=5x+7
Subtract the terms on the left from the terms on the right and you get
3x+2

4)

f(x)= x^2-3x+10

Simply plug -4 in everywhere you see "x"

f(-4)= (-4)^2-3^(-4)+10 = 16+12+10 = 38

Second one is x-5.. Last one is 38

Try mathway.com