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OleMash [197]
3 years ago
5

- 16n}^{5} + {12n}^{3} " alt=" { - 16n}^{5} + {12n}^{3} " align="absmiddle" class="latex-formula">
Please factor out the common factor from the expression​
Mathematics
1 answer:
Lisa [10]3 years ago
8 0

Answer:

-4n^3(4n^2-3)

Step-by-step explanation:

highest common factor of the coefficients are 4

highest common factor of the terms are n^3

hence

-4n^3(4n^2-3)

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A local high school requires at least one adult chaperone for every 12 students on any field trip. The principal uses the inequa
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Which of the following expressions is a factor of the polynomial x2+3/2x-1?
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the answer is x-1/2

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What number do I subtract each time?
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Instructions for a craft project say that the length of a piece of red ribbon should be 7inches than the length of a piece of bl
sergeinik [125]

<em><u>Question:</u></em>

Instructions for a craft project say that the length of a piece of red ribbon should be 7 inches less than the length of a piece of blue ribbon.

How long is the red ribbon if the length of the blue ribbon is:

10 inches?

27 inches?

x inches?

<em><u>Answer:</u></em>

a) Length of red ribbon is 3 inches

b) Length of red ribbon is 20 inches

c) Length of red ribbon is (x - 7) inches

<em><u>Solution:</u></em>

Given that,

The length of a piece of red ribbon should be 7 inches than the length of a piece of blue ribbon

Therefore,

Length of red ribbon = length of blue ribbon - 7 ---- eqn 1

We have to find the length of red ribbon if the length of the blue ribbon is:

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Therefore, from eqn 1,

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Length of red ribbon = 3

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b)  length of blue ribbon = 27 inches

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Length of red ribbon = 27 - 7

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Therefore, from eqn 1,

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4 0
3 years ago
Endpoint P(5, -3) and Q(2,4)
NikAS [45]

The midpoint of these two points would be (3.5, 0.5)

In order to find this, you need to take the average of the two point that you have.  To do that, add the numbers together and divide by 2. Start with just the x values.

(5 + 2)/2 = 3.5

Now the y values

(-3 + 4)/2 = 0.5

This gives us the ordered pair of (3.5, 0.5)

4 0
3 years ago
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