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Deffense [45]
2 years ago
12

Explain how to use a polynomial identity to factor x^6-27y^3

Mathematics
2 answers:
Umnica [9.8K]2 years ago
6 0
Dont really no sorry
Taya2010 [7]2 years ago
5 0

This is a sum of cubes:

<em>x</em>⁶ - 27<em>y</em>³ = (<em>x</em>²)³ + (-3<em>y</em>)³

Recall that

<em>a</em>³ + <em>b</em>³ = (<em>a</em> + <em>b</em>) (<em>a</em>² - <em>ab</em> + <em>b</em>²)

So we have

<em>x</em>⁶ - 27<em>y</em>³ = (<em>x</em>² - 3<em>y</em>) ((<em>x</em>²)² - <em>x</em>² (-3<em>y</em>) + (-3<em>y</em>)²)

<em>x</em>⁶ - 27<em>y</em>³ = (<em>x</em>² - 3<em>y</em>) (<em>x</em>⁴ + 3<em>x</em>²<em>y</em> + 9<em>y</em>²)

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Help me please ASAP
Mashutka [201]

Answer: \frac{5}{35}

Step-by-step explanation:

Since, the total number of contestant = 7,

Thus, the probability that out of 3 contestant, me and my friend is chosen

= \frac{2_C_2\times 5_C_1}{7_C_3}

= \frac{\frac{2!}{2!0!}\times \frac{5!}{1!\times 4!}}{\frac{7!}{4!\times 3!}}

= \frac{1\times 5}{\frac{7\times 6\times 5\times 4!}{4!\times 6}}

=  \frac{5}{35}

⇒ Third Option is correct.

4 0
3 years ago
Is 6/5 closer to 1/2
Alecsey [184]

\frac{6}{5} =\quad 1.2

\frac{1}{2}= \quad 0.5

1.2 > 0.5

\frac{6}{5} > \frac{1}{2}

<h3><em>Hope I helped you!</em></h3><h3><em>Success!</em></h3>
5 0
3 years ago
Read 2 more answers
Evaluate each expression |-16| -|-1|
Maslowich

Understanding the Absolute Value.

First, know what the absolute value is.

The absolute value is the value that determines how far the value is from 0.

For example, The absolute value of -5 is far from 0 5 units. Therefore the absolute value of -5 equals 5.

Basic Absolute Value Defines

| a | = a

- | a | = -a

| - a | = a

Back to the question. To evaluate those expressions, we use the defines of absolute value.

|-16| = 16

|-1| = 1

16-(1)

Then remove the brackets. 16 - 1 = 15

Therefore, the answer is 15.

<em>The</em><em> </em><em>answer</em><em> </em><em>above</em><em> </em><em>is</em><em> </em><em>when</em><em> </em><em>being</em><em> </em><em>subtracted</em><em> </em><em>and</em><em> </em><em>evaluated</em><em> </em><em>from</em><em> </em><em>both</em><em> </em><em>16</em><em>-</em><em>1</em>

<em>Evaluating</em><em> </em><em>for</em><em> </em><em>each</em><em> </em><em>expressions</em><em> </em><em>would</em><em> </em><em>be</em>

<em>|</em><em>-</em><em>16</em><em>|</em><em> </em><em>=</em><em> </em><em>16</em>

<em>-</em><em>|</em><em>-</em><em>1</em><em>|</em><em> </em><em>=</em><em> </em><em>-</em><em>1</em>

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The 7 before the decimal would be classified as a tenth
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