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Sever21 [200]
3 years ago
14

What fraction is between 3/4 and 1 ?

Mathematics
1 answer:
VLD [36.1K]3 years ago
8 0
7/8 is between 3/4 and 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.

3 0
3 years ago
Help Me PLEASE!!! Simplify The Equation (2nd question)
scoray [572]
I think it would be 8-3, not 7. 8x - 2x + 5x = (8-5)x + 5x
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3 years ago
What is the slope of the line that passes through (-4, 5) and (1, 3)?​
Tasya [4]
M=-2/5 is the answer to this equation
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3 years ago
Multiply and and simplify : (a+b)^2<br> Pls help easy
Lena [83]

Answer:a^2+2ab+b^2


Step-by-step explanation:

First write it out in extended form, (a+b)(a+b) then multiply each term by the other in each parenthesis,  such as a*a+b*b+a*b+a*b


Then, once you have done that, simplify it. you will then get the answer a^2+2ab+b^2


3 0
3 years ago
Samantha works at a nursery where she pots plants. It takes her 112 hours to pot 12 plants. Question 1 Part A How many plants ca
PolarNik [594]

Answer:

9

Step-by-step explanation:

112/12= 9.3

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3 years ago
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