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Svetradugi [14.3K]
3 years ago
15

3 times what is 5 (whole number if possible)

Mathematics
2 answers:
valentina_108 [34]3 years ago
8 0

Answer:

1.6, but can be rounded to 2

Step-by-step explanation:

nadezda [96]3 years ago
7 0

Answer:

1.66

Cannot be a whole number because 3 times a whole number cannot equal 5

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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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Lol help anyone?PLEASE !!
Savatey [412]

Answer:

B: 1/5 = 4/x

Step-by-step explanation:

1cm = 5km

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