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2 years ago

6

Sam was asked to place the numbers shown below in order from least to greatest.

1 answer:

Aloiza [94]2 years ago

8 0

Answer:

There's no question here or picture

Step-by-step explanation:

You might be interested in

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

Who can get this correct? Marking most Brainly

Helen [10]

Answer:

I think it's A

Step-by-step explanation:

4 0

2 years ago

If BTS=GHD BS=25 TS=14 BT=31 GD=4x-11 S=56 B=21 and H=(7y+5) find the values of x and y

77julia77 [94]

Given:

Consider the completer question is "If ∆BTS≅∆GHD, BS=25, TS=14, BT=31, GD=4x-11, m∠S=56, m∠B=21 and m∠H=(7y+5), find the values of x and y.

To find:

The values of x and y.

Solution:

We have,

$\Delta BTS\cong \Delta GHD$ (Given)

$BS=GD$ (CPCTC)

$25=4x-11$

$25+11=4x$

$36=4x$

Divide both sides by 4.

$9=x$

In ∆BTS,

$\angle B+\angle T+\angle S=180^\circ$ (Angle sum property)

$21^\circ+\angle T+56^\circ=180^\circ$

$77^\circ+\angle T=180^\circ$

$\angle T=180^\circ-77^\circ$

$\angle T=103^\circ$

Now,

$\angle T=\angle H$ (CPCTC)

$103=7y+5$

$103-5=7y$

$98=7y$

Divide both sides by 7.

$14=y$

Therefore, the value of x is 9 and value of y is 14.

3 0

2 years ago

PLEASE HELP MY ASSIGNMENT IS DUE IN 3 HOURS!!!!

Hatshy [7]

Answer:

76.3 degrees

Step-by-step explanation:

We use inverse cosine to find the angle.

arccos(1.9/8) = 76.26 deg

7 0

2 years ago

Aside from playing scrabble, what ither indoor recreational activities would you intiate in your family/community despite tha pa

olga nikolaevna [1]

Sungka. Playing sungka is fun

6 0

2 years ago