2 years ago

What is the value Of this

Mathematics

1 answer:

Lostsunrise [7]2 years ago

8 0

What’s the question??

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Let n be a natural number. Show that 3 | n3 −n

Vladimir [108]

Let $n=1$ . Then $n^3-n=1^3-1=0$ . By convention, every non-zero integer $n$ divides 0, so $3\vert n^3-n$ .

Suppose this relation holds for $n=k$ , i.e. $3\vert k^3-k$ . We then hope to show it must also hold for $n=k+1$ .

You have

$(k+1)^3-(k+1)=(k^3+3k^2+3k+1)-(k+1)=(k^3-k)+3(k^2+k)$

We assumed that $3\vert k^3-k$ , and it's clear that $3\vert 3(k^2+k)$ because $3(k^2+k)$ is a multiple of 3. This means the remainder upon divides $(k+1)^3-(k+1)$ must be 0, and therefore the relation holds for $n=k+1$ . This proves the statement.

4 0

3 years ago

A customer ordered a product online for $299 and was charged an additional $10 for shipping and handling. The customer received

swat32

Answer:

$192.4

Step-by-step explanation:

First, we'll work out what the customer had to pay before tax.

That's

$299+10-99-25\\=185$ (USD)