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inna [77]
2 years ago
7

Factorise fully p^3 + p^2

Mathematics
1 answer:
nikklg [1K]2 years ago
3 0

Answer:

\huge\boxed{p^3+p^2=p^2(p+1)}

Step-by-step explanation:

p^3+p^2\\\\p^3=p^{2+1}=p^2\cdot p\\\\p^3+p^2=p^2\cdot p+p^2\cdot1=p^2(p+1)

Used:

a^n\cdot a^m=a^{n+m}\\\\\text{distributive property}:\ a(b+c)=ab+ac

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Compute: 245 * 1,493. Show your work.
sesenic [268]
0 + 0 + 5 = 5

0 + 2 + 6 = 8

6 + 7 + 4 = 17
Put the 7 in the appropriate column and carry the 1.

<span>8 + 9 + 7 + </span><span>1(carried)</span><span> = 25</span>
Put the 5 in the appropriate column and carry the 2.

<span>9 + 5 + </span><span>2(carried)</span><span> = 16</span>
Put the 6 in the appropriate column and carry the 1.

<span>2 + </span><span>1(carried)</span><span> = 3</span>

<span>So: 1493 × 245 = 365785</span>
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2 years ago
A basic cellular phone plan costs $4 per month for 70 calling minutes. Additional time costs $0.10 per minute. The formula C= 4+
Harrizon [31]

Answer:

For a monthly cost of at least $7 and at most $8, you can have between 100 and 110 calling minutes.

Step-by-step explanation:

The problem states that the monthly cost of a celular plan is modeled by the following function:

C(x) = 4 + 0.10(x-70)

In which C(x) is the monthly cost and x is the number of calling minutes.

How many calling minutes are needed for a monthly cost of at least $7?

This can be solved by the following inequality:

C(x) \geq 7

4 + 0.10(x - 70) \geq 7

4 + 0.10x - 7 \geq 7

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For a monthly cost of at least $7, you need to have at least 100 calling minutes.

How many calling minutes are needed for a monthly cost of at most 8:

C(x) \leq 8

4 + 0.10(x - 70) \leq 8

4 + 0.10x - 7 \leq 8

0.10x \leq 11

x \leq \frac{11}{0.1}

x \leq 110

For a monthly cost of at most $8, you need to have at most 110 calling minutes.

For a monthly cost of at least $7 and at most $8, you can have between 100 and 110 calling minutes.

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square: 1 1/2, 3 1/2

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7 0
2 years ago
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475=160-(4/5)b is the correct answer
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