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romanna [79]
3 years ago
11

Which two numbers doesStartRoot 61 EndRoot lie between on a number line?

Mathematics
2 answers:
denpristay [2]3 years ago
7 0

Answer:

Between 7 and 8

Step-by-step explanation:

6^2 = 36

7^2 = 49

8^2 = 64

9^2 = 81

since 61 is between 7^2 and 8^2, √61 is between 7 and 8

Romashka [77]3 years ago
3 0

√61 = 7.8102...

7.8102 is in between 7.8 and 7.9

<h3>Answer:</h3>

7.8 and 7.9

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The goal is to find the radius of a circle which also helps to get the diameter. Finding the center helps to measure and know the circumference.

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There were thirteen friends playing a video game online when six players quit.If each player left had eight lives how many lives
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They had 56 lives in total

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If all thirteen friends were still in the game then they would have 104 lives all together.
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3 years ago
Why did I get this question wrong?
Talja [164]

Answer:

9.25

Step-by-step explanation:

M₄ means the area using 4 rectangles with heights equal to the function evaluated at the midpoints.

The width of each rectangle is (4−0)/4 = 1.

The area of each rectangle is:

M₁ = (1) (1.12) = 1.12

M₂ = (1) (1.80) = 1.80

M₃ = (1) (2.69) = 2.69

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7 0
3 years ago
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
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