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Airida [17]
3 years ago
6

Factor x 2 + xy - 2x - 2y. (x + 2)(x - y) (x + y)(x - 2) (x - y)(x + 2)

Mathematics
2 answers:
Y_Kistochka [10]3 years ago
4 0
X² + xy - 2x - 2y
The first thing would be to break the polynomial up into pairs.
(x² + xy) + (-2x -2y)
Then you can factor them individually.
(x² + xy)
Both numbers have x in common, so you can factor it out and get
x(x + y)
Then you factor the other pair,
(-2x -2y)
Both numbers have -2 in them, so you can factor it out and get
-2(x + y)
Now your two pairs are
[x (x + y)] and [-2 (x + y)]
Notice that the two terms in parentheses are the same, so you can get rid of one and combine the two outside terms to get a final answer of
(x - 2)(x + y)
Art [367]3 years ago
4 0

Answer:

Your answer is (x+y)(x-2)

Step-by-step explanation:

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What is the volume of a cone with a radius of 4 centimeters and a height of 9 centimeters? Use 3.14 to estimate p.
lilavasa [31]
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HELP PLEAAASEEE Ireland opened a bag or skittles and recorded the colors and their frequencies.
Anna11 [10]

The correct answer is 25%

<h3>What is Relative Frequency?</h3>
  • The number of times an event occurs divided by the total number of events occurring in a given data.

<h3>How to solve the problem?</h3>
  • This problem can be solved by following steps.
  • The data of color and frequency is given in table.
  • We need find the relative frequency for Purple

First calculate the the total number of frequency

18+20+10+16

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The total number of frequency is 64

Hence the relative frequency of purple is 16/64

Therefore , relative frequency of purple is 1/4 = 0.25

Therefore 25% of purple candies are there

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What are the zeros of the polynomial function: f(x) = x3 – 5x2 -6x
shepuryov [24]

The zeros of the polynomial function: f(x) = x^3 – 5x^2 -6x is -1, 0, 6.

<u>Solution:</u>

Given, polynomial equation is x^{3}-5 x^{2}-6 x

We have to find the zeroes of the given polynomial.

So, let us equate it with 0.

\begin{array}{l}{\text { Then, } x^{3}-5 x^{2}-6 x=0} \\\\ {\rightarrow x\left(x^{2}-5 x-6\right)=0} \\\\ {\rightarrow x\left(x^{2}-(6-1) x-6\right)=0} \\\\ {\rightarrow x\left(x^{2}-6 x+x-6\right)=0} \\\\ {\rightarrow x(x(x-6)+1(x-6))=0} \\\\ {\rightarrow x(x-6)(x+1)=0} \\\\ {\rightarrow x=0 \text { or } x-6=0 \text { or } x+1=0} \\\\ {\rightarrow x=0 \text { or } 6 \text { or }-1}\end{array}

Hence, the roots of the given polynomial are -1, 0, 6.

6 0
3 years ago
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