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3 years ago

How do you simplify 1 π (64)^2 (6)

Mathematics

2 answers:

katrin [286]3 years ago

8 0

Answer:

if you multiply pi by 64^2 times 6 then it comes out to 77207.78

Step-by-step explanation:

64^2 = 4096

4096 (6) = 24576

24576 (π) ≈ 77207.78

erastova [34]3 years ago

3 0

The answer for the question is 24576 π. (Decimal 77207.78015)

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PLEASE HELP WILL GIVE 1ST RIGHT ANSWER AS BRAINIEST

Ne4ueva [31]

Answer:

26π

Step-by-step explanation:

Hmm... This one is a little hard to understand because of the LaTeX.

Any way, way back to the question. A useful piece of information:

<u>The formula for finding the circumference of a circle is 2πr or π · d :</u>

We first need to find out what x is.

Since 2 times the radius is the diameter, we can set up our equation like this:

2(x + 6) = 3x + 5

Solving gives:

2x + 12 = 3x + 5.

We subtract 2x from both sides:

+12 = x + 5

Subtract 5:

So x = 7.

Now we can plug-and-chug:

7 + 6 = 13 times 2pi (this is the radius)

21 + 5 = 26 times pi.

<u>Check:</u>

When we check 13 (radius) times 2 should equal the diameter(26)

13 * 2 = 26.

So we are correct. The answer 26π is correct.

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3 years ago

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Genrish500 [490]

Answer:

Each friend will get 5/8 (excluding Rita)

Each person will get 15/28 (including Rita)

Step-by-step explanation:

You must simply divide 3 3/4 ÷ 6 (excluding Rita)

You must simply divide 3 3/4 ÷ 7 (including Rita)

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3 years ago

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A source of information randomly generates symbols from a four letter alphabet {w, x, y, z }. The probability of each symbol is

koban [17]

The expected length of code for one encoded symbol is

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where $p_\alpha$ is the probability of picking the letter $\alpha$ , and $\ell_\alpha$ is the length of code needed to encode $\alpha$ . $p_\alpha$ is given to us, and we have

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so that we expect a contribution of

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bits to the code per encoded letter. For a string of length $n$ , we would then expect $E[L]=1.375n$ .

By definition of variance, we have

$\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2$

For a string consisting of one letter, we have

$\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha{\ell_\alpha}^2=\dfrac12+\dfrac{2^2}4+\dfrac{2\cdot3^2}8=\dfrac{15}4$

so that the variance for the length such a string is

$\dfrac{15}4-\left(\dfrac{11}8\right)^2=\dfrac{119}{64}\approx1.859$

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den301095 [7]

Answer:

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3 years ago