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stiks02 [169]
3 years ago
6

Prove that (4n+1)^2(4n-1)^2 is a multiple of 8 for all integers of n. thanks.

Mathematics
1 answer:
azamat3 years ago
6 0

Answer: i think  (4n+1)^2(4n-1)^2 isnt a multiple of 8 for all integers of n because:

(4n + 1)²(4n - 1)²

= [(4n + 1)(4n - 1)]²

= (16n² - 1)²

= 16².n².n² - 2.16.n² + 1

= 8n²(32n² - 4) + 1

can see 8n²(32n² - 4) is a multiple of 8 but 1 isnt a multiple of 8

=> (4n + 1)²(4n - 1)² isnt a multiple of 8 for all integers of n.

Step-by-step explanation:

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Answer:

24

Step-by-step explanation:

We know that the final answer has to be in inches.

To convert ft = in, multiply given ft. by 12.

1/6 = 1/6 * 12 = 2

1/4 = 0.25 * 12 = 3

1/3 = 0.3 * 12 = 4

2*3*4 = 23 cubic in.

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1 year ago
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a bag contains 5 blue, 8 red, and 7 green marbles. a marble is selected at random. Write each answer as a fraction, percent, and
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Answer:

1/4 and 0.25 and 25% chance for blue, 2/5 and 0.4 and 40% chance for red, and 7/20 and 0.35 and 35% for green.

Step-by-step explanation:

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If f(p) divided by x-p and x-q have the same remainder
oee [108]
Hello,


x^2-y^2=(x+y)(x-y)
x^3-y^3=(x-y)(x²+xy+y²)

Let's use Horner's division

.........|a^3|a^2.|a^1..........|a^0
.........|1....|5....|6..............|8....
a=p...|......|p....|5p+p^2....|6p+5p^2+p^3
----------------------------------------------------------
.........|1....|5+p|6+5p+p^2|8+6p+5p^2+p^3

The remainder is 8+6p+5p^2+p^3 or 8+6q+5q^2+q^3


Thus:
8+6p+5p^2+p^3 = 8+6q+5q^2+q^3
==>p^3-q^3+5p^2-5q^2+6p-6p=0
==>(p-q)(p²+pq+q²)+5(p-q)(p+q)+6(p-q)=0
==>(p-q)[p²+pq+q²+5p+5q+6]=0 or p≠q
==>p²+pq+q²+5p+5q+6=0

And here, Mehek are there sufficients explanations?
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Answer:

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Explanation:

I used a calculator for the volume online

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