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lutik1710 [3]
3 years ago
8

Explain how you could use the information that all circles are similar to show that the value of pi is a constant.

Mathematics
1 answer:
zhenek [66]3 years ago
7 0

Answer:

The proof that πk(C1)=πk(C2) of course would just apply the similarity of polygons and the behavior of length and area for changes of scale. This argument does not assume a limit-based theory of length and area, because the theory of length and area for polygons in Euclidean geometry only requires dissections and rigid motions ("cut-and-paste equivalence" or equidecomposability). Any polygonal arc or region can be standardized to an interval or square by a finite number of (area and length preserving) cut-and-paste dissections. Numerical calculations involving the πk, such as ratios of particular lengths or areas, can be understood either as applying to equidecomposability classes of polygons, or the standardizations. In both interpretations, due to the similitude, the results will be the same for C1 and C2.

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Write 120 as a product of prime numbers
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Here is the process:

120/ 2 = 60

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30 / 2 = 15

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Help!! Simplify (see photo) pls explain.
wolverine [178]

Answer:

The answer to your question is: 4x²y³

Step-by-step explanation:

Simplify                                     ∛(-64x⁶y⁹)

                                                 

First, find the prime factors of 64

      64   2

      32   2

      16    2

       8    2

       4    2

       2    2

        1              then 64 = 2⁶

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Roots can be expressed as fractional powers where the numerator is the power of the number inside the root and the denominator is the root.

Ex.        -2 ⁶/³       x⁶/³      y⁹/³

Now simplify

                       (-2)²  x²  y ³

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3 years ago
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san4es73 [151]

Answer:

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Step-by-step explanation:

Considering:

$ (x+y)^n = \sum_{k=0}^{n} \binom{n}{k}  x^{n-k}y^k$

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Now, you gotta calculate for every value of k

$ (2x-3y)^6 = \binom{6}{0}  (2x)^{6-0}(-3y)^0     +       \binom{6}{1}  (2x)^{6-1}(-3y)^1     +      \binom{6}{2}  (2x)^{6-2}(-3y)^2   +   \\ $

$\binom{6}{3}  (2x)^{6-3}(-3y)^3    +    \binom{6}{4}  (2x)^{6-4}(-3y)^4    +  \binom{6}{5}  (2x)^{6-5}(-3y)^5    +    \binom{6}{6}  (2x)^{6-6}(-3y)^6            $

I will not write every product, but just solve following the steps:

For k=0

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$1\cdot \:1\cdot \left(2x\right)^{6-0}$

$2^6x^6$

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melamori03 [73]

Answer:

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Step-by-step explanation:

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\sqrt{14}

5 0
4 years ago
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