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Nookie1986 [14]

3 years ago

11

PLEASE ANSWER 16 17 18 CORRECTLY I WILL MARK BRAINLIST

1 answer:

nadya68 [22]3 years ago

8 0

Answer:

A A B

Step-by-step explanation:

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Deffense [45]

Answer:

I think it is c correct me if I'm wrong

6 0

3 years ago

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Which measure of center is the most appropriate for this

belka [17]

12. If this isn't right, please tell me and I will fix it.

Step-by-step explanation:

4 0

3 years ago

What is and how do you do it?<br><br> H(x) = 4x-2<br> G(x) = x^2 -5<br><br> Find h(x) + g(x)

sineoko [7]

$h(x)=4x-2\\\\(x)=x^2-5\\\\h(x)+g(x)=(4x-2)+(x^2-5)=x^2+4x+(-2-5)\\\\=x^2+4x-7$

6 0

3 years ago

Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$

Viktor [21]

When $n=0$ , we have

$5^{3^0} + 1 = 5^1 + 1 = 6$

$3^{0 + 1} = 3^1 = 3$

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for $n=k$ , that

$3^{k + 1} \mid 5^{3^k} + 1$

Now for $n=k+1$ , we have

$5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)$

so we know the left side is at least divisible by $3^{k+1}$ by our assumption.

It remains to show that

$3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1$

which is easily done with Fermat's little theorem. It says

$a^p \equiv a \pmod p$

where $p$ is prime and $a$ is any integer. Then for any positive integer $x$ ,

$5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3$

Furthermore,

$5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3$

which goes all the way down to

$5^{3^k} \equiv 5 \pmod 3$

So, we find that

$\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3$

QED

5 0

2 years ago

The value of a is the value of B is the value of C is

enyata [817]

Answer:

A: 5

B: 9

C: 32

I hope this helps the people who're looking for this answer!

5 0

3 years ago