Perform the indicated operation and simply the result.

2 answers:

5 0

I am assuming that the whole 3x - 6 in the fraction is the numerator

the this simplifies to 3x (3x - 6)

= 9x^2 - 18x (C)

Murrr4er [49]3 years ago

4 0

Answer:

9x^2 - 18x

Hope this helps! :)

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Factor 3x^2y^2 − 2xy^2 − 8y^2. Show your work.

docker41 [41]

3x^2y^2 − 2xy^2 − 8y^2 =
y^2 (3x^2 - 2x - 8) =

factoring with leading coefficient:

for ax2+bx+c find two numbers n,m, that m*n = a*c and m+n = b

3x^2 - 2x - 8
a=3, b=-2, c=-8
a*c = 3*(-8) = -24
-24=(-6)*4 and -6+4=-2, so m=-6 and n=4

replace bx with mx + nx and factor by grouping

3x^2 - 2x - 8 = 3x^2 -6x + 4x -8 = 3x(x-2) + 4(x-2) = (3x+4)(x-2)

answer:
3x^2y^2 − 2xy^2 − 8y^2 = y^2(3x+4)(x-2)

7 0

3 years ago

Mike had 119 dollars to spend on 6 books. after buying them he had 11 dollars. how much did each book cost ?

deff fn [24]

18 $ Hope I helped
This is how I came to the result..
119-11= 108
108 : 6 = 18 $

8 0

2 years ago

Marta is solving the equation S = 2Πrh + 2Πr2 for h. What is the result?

yulyashka [42]

S = 2Πrh + 2Πr2
Manipulating the equation for h.
Step 1. subtract 2Πr2 on both sides.

S - 2Πr2 = 2Πrh + 2Πr2 - 2Πr2
S - 2Πr2 = 2Πrh

Step 2 . divide 2Πr on both sides

(S - 2Πr2)/2Πr = 2Πrh/2Πr

h = (S - 2Πr2)/2Πr

6 0

3 years ago

A circle has the equation 2x²+12x+2y²−16y−150=0.

KonstantinChe [14]

Answer: B. The coordinates of the center are (-3,4), and the length of the radius is 10 units.

Step-by-step explanation:

The equation of a circle in the center-radius form is:

$(x-h)^{2} +(y-k)^{2}=r^{2}$ (1)

Where $(h,k)$ are the coordinates of the center and $r$ is the radius.

Now, we are given the equation of this circle as follows:

$2x^{2}+12x+2y^{2}-16y-150=0$ (2)

And we have to write it in the format of equation (1). So, let's begin by applying common factor 2 in the left side of the equation:

$2(x^{2}+6x+y^{2}-8y-75)=0$ (3)

Rearranging the equation:

$x^{2}+6x+y^{2}-8y=75$ (4)

$(x^{2}+6x)+(y^{2}-8y)=75$ (5)

Now we have to complete the square in both parenthesis, in order to have a perfect square trinomial in the form of $(a\pm b)^{2}=a^{2}\pm+2ab+b^{2}$ :