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irina [24]
2 years ago
13

Find The Surface Area Of The Pyramid

Mathematics
2 answers:
Travka [436]2 years ago
7 0
75
the formula is
=12pl+B
Salsk061 [2.6K]2 years ago
7 0
Answer is 75 i hope this helps you
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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
Whats the slope of (2,-5),(-6,5)​
Makovka662 [10]
The slope of the line with the points (2,-5) and (-6,5) is 5/-4 (or 10/-8).
3 0
3 years ago
Can u please help me please. Please
jonny [76]
The answer is
C = 4 ÷ 1/5
6 0
3 years ago
This has two answers so ya give um to me
alexandr1967 [171]

Let's separate this absolute value equation into two different equations.

x + 1 = 3.

Subtract 1 from each side

x = 2.

x + 1 = -3

Subtract 1 from each side

x = -4

The two solutions are x = 2 and x = -4

5 0
3 years ago
Find the equation of this linear model
kompoz [17]

Answer:

Is there a diagram to this question?

Step-by-step explanation:

7 0
2 years ago
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