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JulijaS [17]
3 years ago
11

Need help now!!! 20 points!!!

Mathematics
2 answers:
Archy [21]3 years ago
8 0

- x {}^{2}  + x - 1
The answer would be C
s344n2d4d5 [400]3 years ago
8 0

- {x}^{2} + x - 1
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Use induction to prove: For every integer n > 1, the number n5 - n is a multiple of 5.
nignag [31]

Answer:

we need to prove : for every integer n>1, the number n^{5}-n is a multiple of 5.

1) check divisibility for n=1, f(1)=(1)^{5}-1=0  (divisible)

2) Assume that f(k) is divisible by 5, f(k)=(k)^{5}-k

3) Induction,

f(k+1)=(k+1)^{5}-(k+1)

=(k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1)-k-1

=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k

Now, f(k+1)-f(k)

f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-(k^{5}-k)

f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-k^{5}+k

f(k+1)-f(k)=5k^{4}+10k^{3}+10k^{2}+5k

Take out the common factor,

f(k+1)-f(k)=5(k^{4}+2k^{3}+2k^{2}+k)      (divisible by 5)

add both the sides by f(k)

f(k+1)=f(k)+5(k^{4}+2k^{3}+2k^{2}+k)

We have proved that difference between f(k+1) and f(k) is divisible by 5.

so, our assumption in step 2 is correct.

Since f(k) is divisible by 5, then f(k+1) must be divisible by 5 since we are taking the sum of 2 terms that are divisible by 5.

Therefore, for every integer n>1, the number n^{5}-n is a multiple of 5.

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3 years ago
the seventh grade class is raising money to have a class trip at the end of the year they began the year with $1500 now they hav
Novay_Z [31]

Answer:

10%

Step-by-step explanation:

(1650 - 1500) /1500 = 10%

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Lin rode her bike 2 miles in 8 minutes. She rode at a constant speed. How
hodyreva [135]

Answer: 4

Step-by-step explanation:

For this question, we have to divide the number 8 by 2.

8 ÷ 2 = 4

Now, we have to take the quotient, 4, and multiply it by 1.

1 x 4 = 4

The answer is 4.

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