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shutvik [7]
3 years ago
7

Write an equation for the polynomial graphed below y(x)=_________

Mathematics
1 answer:
polet [3.4K]3 years ago
5 0

Answer:

<u>dont know the answer i feel sry for u</u>

:    <u>)</u>

<u />

Step-by-step explanation:

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If a1=7 and an=an-1+5 than find the value of a5<br><br> ​<br> .
Katyanochek1 [597]

Answer:

a₅=27.

Step-by-step explanation:

1) according to the condition the given subsequence is arithmetic sequence. It means, the a₁=7 and d=5, then

2) a₅=a₁+4*d; => a₅=7+4*5=27.

4 0
2 years ago
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
erma4kov [3.2K]

Answer:

\frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}

Step-by-step explanation:

Given

5 tuples implies that:

n = 5

(h,i,j,k,m) implies that:

r = 5

Required

How many 5-tuples of integers (h, i, j, k,m) are there such thatn\ge h\ge i\ge j\ge k\ge m\ge 1

From the question, the order of the integers h, i, j, k and m does not matter. This implies that, we make use of combination to solve this problem.

Also considering that repetition is allowed:  This implies that, a number can be repeated in more than 1 location

So, there are n + 4 items to make selection from

The selection becomes:

^{n}C_r => ^{n + 4}C_5

^{n + 4}C_5 = \frac{(n+4)!}{(n+4-5)!5!}

^{n + 4}C_5 = \frac{(n+4)!}{(n-1)!5!}

Expand the numerator

^{n + 4}C_5 = \frac{(n+4)!(n+3)*(n+2)*(n+1)*n*(n-1)!}{(n-1)!5!}

^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5!}

^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5*4*3*2*1}

^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}

<u><em>Solved</em></u>

6 0
3 years ago
If a pyramid has a square base with side length s and height h, which formula represents the volume of the pyramid?
notsponge [240]
V = s^2(h/3)
is the formula....................<span />
8 0
3 years ago
Read 2 more answers
Imagine the United States only had two postage stamps, a 3 cent stamp and a 7 cent stamp. If you can put any number of stamps on
cricket20 [7]

Answer:

  • not possible:  1, 2, 4, 5, 8, 11
  • possible: all others

Step-by-step explanation:

Obviously, all multiples of 3 and 7 can be paid.

1 cent can be added by adding a 7-cent stamp and taking away two 3-cent stamps. 2 cents can be added by adding three 3-cent stamps and taking away one 7-cent stamp. Hence any value more than 7 +2(3) = 13 cents can always be accommodated.

Amounts that are impossible:

  1, 2, 4, 5, 8, 11 cents

All other positive integer amounts are possible.

7 0
3 years ago
Help me answer this question please
valentina_108 [34]
The median is 10.5 because its the middle line in the box and all you have to do is look at it and you can see it
8 0
3 years ago
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