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zzz [600]
3 years ago
12

Find the length of the radius of the circle.

Mathematics
2 answers:
Kruka [31]3 years ago
7 0

Answer:

By Pythagoras theorem, we get

{x}^{2}  +  {x}^{2}  =  {( \sqrt{72}) }^{2} \\ 2 {x}^{2}  =  72 \\  {x}^{2}  = 36 \\  \boxed{x = 6}

<h2>C) <u>6 inches</u> is the right answer.</h2>
bearhunter [10]3 years ago
3 0

Answer:

C)  6 inches

Step-by-step explanation:

(√72)² = x² + x²

72 = 2x²

72/2 = x²

36 = x²

√36 = √x²

6 = x

x = 6 inches

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Estimate the sum of 279 and 217
Shalnov [3]
280 + 220 = 500 is the answer
6 0
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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
(Kinda need help ASAP)
denis23 [38]

Answer:

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4 0
3 years ago
An equation in the form ax2+bx+c=0 is solved by the quadratic formula. The solution to the equation is shown below.x=−7±√ "572"
7nadin3 [17]

Answer:

B: a = 1, b= 7, c = -2

Explanation:

\sf Quadratic \ Equation : \dfrac{-b\pm \sqrt{b^2 - 4ac} }{2a}

Knowing This:

<u>solve for b</u>

  • -b = -7
  • b = 7

<u>solve for a</u>:

  • 2a = 2
  • a = 1

<u>solve c</u>:

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  • 7² - 4(1)(c) = 57
  • -4c = 57 - 49
  • -4c = 8
  • c = 8/-4 = -2
4 0
2 years ago
Read 2 more answers
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