Prove that there is a prime between n and n factorial

1 answer:

max2010maxim [7]3 years ago

8 0

Let k = n! - 1. Well, since all numbers from 2 to n divide n!, none of these divides k. This means that k either a prime , in which case we are done, or there exists a prime p > n which divides n! - 1. In the latter case, p is between n and n factorial and p is prime, so the proof is complete.

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Please help. Will give brainliest. And I need step by step please

Elan Coil [88]

Answer: 500cm squared

Step-by-step explanation:

To figure out the area, here are the steps

First, seperate the parallelogram into 2 triangles and a rectangle.

Secondly, we see it has figured out the area of the triangle which is 25, so we can assume that the other triangle is also 25. So the area of the two triangles are 25 + 25 = 50

Then, if we look at what it has labeled, we would know that the length of the parallelogram is 50, BUT this 50 contains ONE length of the triangle, which means to get the actual length of the rectangle, we must subtract a length of a triangle from 50.

That's 50 - 5 = 45. Therefore the area of the rectangle is (length times width) 45 ✕ width = the area of the rectangle.

Now we realize that we don't know the width of the rectangle, we can figure that out by back-tracking from the area of the triangle.

We know that area of a triangle is length times height divided by 2, so 25 ✕ 2 divided by legth would get us the height of the triangle. 25 ✕ 2 ÷ 5 = 10

Now we can figure out the area of the rectangle. 45 ✕ 10 = 450

Add the rectangle and the triangles together, you get -- 450 + 50 = 500 (cm squared）

7 0

3 years ago

Read 2 more answers

What's the equation for this problem? Please help!! It's due tomarrow

prohojiy [21]

Wth diaiqowjqksosjshbssbbsjsoaiajw

5 0

3 years ago

Prove whether the following are identities 2tanh 1/2x / 1−tanh^2 1/2 x = sinh x

Tanzania [10]

Recall that

$\cosh^2(x) - \sinh^2(x) = 1$

Dividing both sides by cosh²(x) gives

$1 - \tanh^2(x) = \mathrm{sech}^2(x)$

Also, recall the identity

$\sinh(2x) = 2\sinh(x)\cosh(x)$

Then

$\dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = \dfrac{2\tanh\left(\frac x2\right)}{\mathrm{sech}^2\left(\frac x2\right)} \\\\ \dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = 2\tanh\left(\dfrac x2\right)\cosh^2\left(\dfrac x2\right) \\\\ \dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = 2\sinh\left(\dfrac x2\right)\cosh\left(\dfrac x2\right) \\\\\dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = \sinh(x)$