Hello from MrBillDoesMath!
Answer:
(x+2) ( x^2 + 3)
Discussion:
x^3 + 2x^2 + 3x + 6 =
(
x^3 + 2x^2) + (3x + 6) = => factor "x^2" from the first term; factor 3
from the second term
x^2( x + 2) + 3( x+2) = => factor (x+2) from each term
(x+2) ( x^2 + 3)
Thank you,
MrB
Answer:
(1/2), (3/2)
Step-by-step explanation:
midpoint fomula
(x1+x2)/ 2, (y1+y2)/2
(5-4)= 1 1/2 = (1/2)
(-1+4)=3 3/2=(3/2)
If x=-6 then y=54.
If x=-4 then y=51.
If x=-2 then y=48.
If x=0 then y=45.
If x=2 then y=42.
If x=4 then y=39.
If x=6 then y=36.
There are an infinite variety of pairs of numbers for 'x' and 'y'
that can make that equation true.
You can use that equation to draw a line on a graph. Then
EVERY point on the line is a solution to the equation.
'x' and 'y' don't have single values unless you have TWO equations.
Answer:
4.8
Step-by-step explanation:
Um, I'm not sure how to explain, but watch a khan academy video on it
At the end of the year, Juan has 52.71 more than 4 times his balance at the beginning. Okay, let's set this up.
4x + 52.71
(4 times) (52.71 more)
His ending was 172.90, so
4x + 52.71=172.90
4x= 120.19
x= 30.05
He had $30.05 at the beginning of the year.
