2 years ago

Which Figure is a rotation of the Blue Figure? A B C D

Mathematics

1 answer:

alexdok [17]2 years ago

7 0

B is the rotation of the Blue figure

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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

Given: *ABC ~ *DMV ; DM = ___

jeka57 [31]

Answer:

DM should equal 20 inches

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2 years ago

How many solutions does 3 - 2x = 5 - x + 3 + 4x have?

vitfil [10]

Well let's see.

$3-2x=5-x+3+4x \\3-2x=8+3x \\-5x=5 \\x=-1 \\$

The equation has exactly one solution.

Hope this helps.

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2 years ago

To rewrite the expression ab+a+b+1, Jozelle first writes (ab+a) + (b+1). What property justifies this?

rodikova [14]

Answer:

I believe it is associative property of addition or commutative

Step-by-step explanation:

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2 years ago