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Temka [501]
3 years ago
5

Find the area of the Compound shape The answer is 89 m^2 I just need the steps....

Mathematics
1 answer:
Lera25 [3.4K]3 years ago
7 0

Answer:

Step-by-step explanation:

Area of semi-circle is \frac{\pi r^2}{2} ≈ \frac{22}{7} × 4² ÷ 2 ≈ 25 cm²

Area of square is 8² = 64 cm²

Total area is 25 + 64 = 89 cm²

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3 years ago
Solve the right triangle. Round decimal answers to the nearest tenth. find the sides and angles
babymother [125]

Answer:

b = 12.6

Step-by-step explanation:

a*2 + b*2 = c*2

6*2 + b*2 = 14*2

36 + b*2 = 196

Subtract 36 from both sides

36 + b*2 = 196

-36             -36

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Each batch of granola Lena makes uses 3 cups of oats, 1 cup of raisins, and 2 cups of nuts. Lena wants to make 4 batches of gran
pogonyaev

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4 0
3 years ago
Read 2 more answers
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
Can someone help me with math
kkurt [141]
A is the correct answer, I believe.
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3 years ago
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