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3 years ago

PLEASE HELP I'M DESPERATE!

Mathematics

1 answer:

Anettt [7]3 years ago

4 0

The upper comment is correct

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(5,-6) after a rotation of 90 digres

vlada-n [284]

Answer:

depends which way you rotate it

Step-by-step explanation:

counterclockwise rotation: (6,5)

clockwise rotation: (-6,-5)

7 0

3 years ago

Help pls its a test t t t t t t

andrew-mc [135]

Answer: C

Step-by-step explanation: Hope this help :D

8 0

3 years ago

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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

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

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

What is the constant in this algebraic expression? 7d+2 I need help

snow_lady [41]

It is 2 because no variable

6 0

3 years ago

A plant costs p dollars and a bush costs b dollars. Ana buys 2 plants and 4 bushes for $42. Paola buys 7 plants and 9 bushes for

gregori [183]

Answer:

p = 5 and b = 8

Step-by-step explanation:

First, create a system of equations:

2p + 4b = 42

7p + 9b = 107

Solve by elimination by multiplying the top equation by -7 and the bottom equation by 2:

-14p - 28b = -294

14p + 18b = 214

Add these together, and solve for b:

-10b = -80

b = 8

Plug in 8 as b into one of the equations, and solve for p:

2p + 4b = 42

2p + 4(8) = 42

2p + 32 = 42

2p = 10

p = 5

So, p = 5 and b = 8

4 0

3 years ago

Read 2 more answers