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schepotkina [342]
3 years ago
8

Reflect the shape against the line x=4

Mathematics
1 answer:
OLga [1]3 years ago
6 0

Answer: Reflect accross the gradient x=4

Step-by-step explanation: Make sure one of the lines or whatever you are drawing is touching or passing thorugh the gradient x = 4

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Please calculate this limit <br>please help me​
Tasya [4]

Answer:

We want to find:

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}

Here we can use Stirling's approximation, which says that for large values of n, we get:

n! = \sqrt{2*\pi*n} *(\frac{n}{e} )^n

Because here we are taking the limit when n tends to infinity, we can use this approximation.

Then we get.

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} =  \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}

Now we can just simplify this, so we get:

\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\

And we can rewrite it as:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n}

The important part here is the exponent, as n tends to infinite, the exponent tends to zero.

Thus:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n} = \frac{1}{e}*1 = \frac{1}{e}

7 0
3 years ago
What’s he answer thanks
gayaneshka [121]

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Answer:

B.) StartFraction 5 StartRoot 11 EndRoot + 5 StartRoot 3 EndRoot Over 8 EndFraction

Step-by-step explanation:

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wel
The car would travel 135 miles bc you multiply 81×5 and get 405 ,then divide that answer by 3
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Consider the following equation. cos x = x3 (a) Prove that the equation has at least one real root. f(x) = cos x − x3 is continu
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