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valentina_108 [34]
2 years ago
15

Describe the end behavior of the following function f(x)=-7x^3+2x-4

Mathematics
2 answers:
Nat2105 [25]2 years ago
8 0
(-0,4),,,,;,,,,,,,,,,,,,,,,,,,
GaryK [48]2 years ago
3 0

Answer:

(0,-4)

Step-by-step explanation:

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Tyson bought four roses for $8.88. If the cost per rose is constant, how much would 6 roses cost?
jeka94

Answer:

6 roses = $13.32

Step-by-step explanation:

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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

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

QED

5 0
1 year ago
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oee [108]
Factor by grouping
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2 p^3+6 p^2+3 p+9=\left(2 p^3+6 p^2\right)+(3 p+9)=\left(2 p^2\right) (p+3)+3 (p+3)
2 p^2 (p+3)+3 (p+3)
(p+3) \left(2 p^2+3\right)
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3 years ago
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Option C- Angelica's side lengths were too long.
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