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2 years ago

Lewis walked 9/10 of a mile. His friend walked 2/5 of the way with him. How many miles did Lewis's friend walk with him?

Mathematics

1 answer:

iVinArrow [24]2 years ago

5 0

Answer:

Lewis's friend walked with him 9/25 miles

Step-by-step explanation:

Given that

Lewis walked 9/10 of a mile

And His friend accompanied him on 2/5 of the way.

This is a simple question of fraction multiplication

The fraction 2/5 has to multiplied with the miles walked by lewis

$= \frac{2}{5} * \frac{9}{10}\\=\frac{9}{25}$

Hence,

Lewis's friend walked with him 9/25 miles

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A. You would divide it between the two

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2 years ago

the angle of depression from an airplane at an altitude of 8000 feet to the airport is 7 degrees. Find the direct distance from

Aleks [24]

Answer:

direct distance = 65,644.07 feet

horizontal distance = 65,154.77 feet

Step-by-step explanation:

The angle of depression is the angle that the airplane does with the horizontal plane.

So, if this angle is 7° and the airplane is at an altitude if 8000 feet, we can find the direct distance and the horizontal distance using the tangent and the sine relations of the angle.

In the tangent of the angle, the opposite side will be the height of the airplane, and the adjacent side will be the horizontal distance (hd), so:

tangent(7) = 8000 / hd

hd = 8000 / tangent(7) = 65,154.77 feet

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sine(7) = 8000 / d

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3 years ago

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

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1 year ago

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3 years ago

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3 years ago