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Mathematics

2 answers:

frozen [14]3 years ago

8 0

Answer:

Step-by-step explanation:

Vlada [557]3 years ago

7 0

The last one is the right answer

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Please answer the 2 question <br> l will mark as brainiest

N76 [4]

Answer:

1. A=πr2=π·4.62≈66.4761

A≈66.48

2. A=πr2=π·22≈12.56637

A≈12.57

please mark as brainiest

3 0

3 years ago

Read 2 more answers

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

211 times 9=<—round the answer.

Yuki888 [10]

Answer:

211 × 9 is 1899 so rounding to nearest thousands would be 2000

3 0

3 years ago

Please help!

IgorC [24]

Answer:

(1,-4) is in quadrant IV.