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3 years ago

If the following is a monomial, write it in standard form.

Mathematics

1 answer:

Sati [7]3 years ago

8 0

Answer:

Monomial; Standard form: 18a^2b^2

Step-by-step explanation:

$6 {a}^{2} 3 {b}^{2} = 6 \times 3 {a}^{2} {b}^{2} = 18{a}^{2} {b}^{2} \\$

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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

1 year ago

What is the vertex of the graph of the function below y=x^2+10x+24

swat32

The vertex is (-5,-1)

7 0

3 years ago

Read 2 more answers

Can someone help me with this

scoray [572]

Answer:

Step-by-step explanation:

7 0

2 years ago

Ruby stands at 5 feet tall her shadow is 6 feet long how far is it form the top of her head to the end of the shadow

alisha [4.7K]

Answer: 7.81 feet

Step-by-step explanation:

This scenario forms a right-angled triangle where Ruby's height of 5 feet is the height of the triangle. Her 6 feet long shadow is the length of the triangle and the distance from the top of her head to the end of the shadow is the hypotenuse.

This can therefore be solved by the Pythagorean formula:

c² = a² + b²

where ;

c = hypotenuse

a = height

b = length

c² = 5² + 6²

c² = 25 + 36

c² = 61

c = √61

c = 7.81 feet

7 0

3 years ago

The following are the temperatures in °C for the first 10 days in January:

Lady bird [3.3K]

Answer:

15.4 is the range.

Step-by-step explanation:

8 0

2 years ago