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1 year ago

If y varies inversely with x and y=2 when x=5, what is the value of x when y=3?

Mathematics

1 answer:

Nostrana [21]1 year ago

8 0

Answer:

Step-by-step explanation:

Method

Inversely in this sentence means

y = k/x

So you first step is to find the value of k. You do that by setting up an inverse relationship with y = 2 , x = 5

2 = k/5 Multiply both sides by 5

2*5 = 5*k/5

10 = k

Now you are able to answer the question.

k = 10
y = 3
x = ?

y = 10/x

3= 10/x Multiply both sides by x

3*x = x * 10/x

3x = 10 Divide both sides by 3

3x/3 = 10/3

x = 3.33 or x = 3 1/3

Answer

x = 3.33 or x = 3 1/3

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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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2 years ago

Pls, can someone help me with this problem?

ExtremeBDS [4]

Answer:

$0.\overline{33}$

Step-by-step explanation:

If you divide number 1 by number 3, you will notice that number 3 will constantly repeat at the decimal part.

So, the answer is $0.\overline{33}$ .

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