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Julli [10]
2 years ago
13

There is a constant ratio between corresponding sides of similar figures. Since circles don't have sides, what ratio(s) could yo

u use to prove that all circles are similar?
Mathematics
2 answers:
inessss [21]2 years ago
7 0
A circle's size and shape is fully defined by its radius. Given two circles with radii r and r', the diameters are d=2r and d'=2r' and they are in the ratio 

<span>d'/d = (2r')/(2r) = r'/r. </span>

<span>The diameter ratio is the same as the radius ratio. Similarly, the circumferences c=πd and c' = πd' are in the ratio </span>

<span>c'/c = (πd')/(πd) = d'/d = r'/r </span>

<span>The circumference ratio is the same as the diameter ratio and the radius ratio. All of the key linear dimensions are in the same proportion. </span>

<span>You might point out that the same thing happens with a square, where the size and shape are also completely determined by a single measurement, the length s of a side, with the diagonal and perimeter (corresponding to diameter and circumference) being d = √2 s and p = 4s. </span>

<span>Maybe you can lead at least some of the students to generalize to other regular polygons. Some of them (like the equilateral triangle and regular hexagon) can be demonstrated like the square and circle above with formulas from geometry. The general case needs trig ratios to state the formulas relating side length to the radius and apothem of the polygon.</span>
hodyreva [135]2 years ago
5 0

Answer:

The ratio of circumference to diameter in all circles is a constant value, pi. Also, the ratio of the diameters and the ratio of the circumferences of any two circles reduces to the ratio of their radii.

Step-by-step explanation:

just answered question, answer above is what it said to be a possible answer

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What is the factorization of 216x12 – 64?
GalinKa [24]

9514 1404 393

Answer:

  (c)  (6x^4 – 4)(36x^8 + 24x^4 + 16)

Step-by-step explanation:

The correct factoring can be found by looking at the first exponent in the second set of parentheses.

The factorization of ...

  (a^3 -b^3)

is ...

  (a -b)(a^2 +ab +b^2)

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Compute the sum:
Nady [450]
You could use perturbation method to calculate this sum. Let's start from:

S_n=\sum\limits_{k=0}^nk!\\\\\\\(1)\qquad\boxed{S_{n+1}=S_n+(n+1)!}

On the other hand, we have:

S_{n+1}=\sum\limits_{k=0}^{n+1}k!=0!+\sum\limits_{k=1}^{n+1}k!=1+\sum\limits_{k=1}^{n+1}k!=1+\sum\limits_{k=0}^{n}(k+1)!=\\\\\\=1+\sum\limits_{k=0}^{n}k!(k+1)=1+\sum\limits_{k=0}^{n}(k\cdot k!+k!)=1+\sum\limits_{k=0}^{n}k\cdot k!+\sum\limits_{k=0}^{n}k!\\\\\\(2)\qquad \boxed{S_{n+1}=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n}

So from (1) and (2) we have:

\begin{cases}S_{n+1}=S_n+(n+1)!\\\\S_{n+1}=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n\end{cases}\\\\\\&#10;S_n+(n+1)!=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n\\\\\\&#10;(\star)\qquad\boxed{\sum\limits_{k=0}^{n}k\cdot k!=(n+1)!-1}

Now, let's try to calculate sum \sum\limits_{k=0}^{n}k\cdot k!, but this time we use perturbation method.

S_n=\sum\limits_{k=0}^nk\cdot k!\\\\\\&#10;\boxed{S_{n+1}=S_n+(n+1)(n+1)!}\\\\\\&#10;

but:

S_{n+1}=\sum\limits_{k=0}^{n+1}k\cdot k!=0\cdot0!+\sum\limits_{k=1}^{n+1}k\cdot k!=0+\sum\limits_{k=0}^{n}(k+1)(k+1)!=\\\\\\=&#10;\sum\limits_{k=0}^{n}(k+1)(k+1)k!=\sum\limits_{k=0}^{n}(k^2+2k+1)k!=\\\\\\=&#10;\sum\limits_{k=0}^{n}\left[(k^2+1)k!+2k\cdot k!\right]=\sum\limits_{k=0}^{n}(k^2+1)k!+\sum\limits_{k=0}^n2k\cdot k!=\\\\\\=\sum\limits_{k=0}^{n}(k^2+1)k!+2\sum\limits_{k=0}^nk\cdot k!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\&#10;\boxed{S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n}

When we join both equation there will be:

\begin{cases}S_{n+1}=S_n+(n+1)(n+1)!\\\\S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\end{cases}\\\\\\&#10;S_n+(n+1)(n+1)!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\\\&#10;\sum\limits_{k=0}^{n}(k^2+1)k!=S_n-2S_n+(n+1)(n+1)!=(n+1)(n+1)!-S_n=\\\\\\=&#10;(n+1)(n+1)!-\sum\limits_{k=0}^nk\cdot k!\stackrel{(\star)}{=}(n+1)(n+1)!-[(n+1)!-1]=\\\\\\=(n+1)(n+1)!-(n+1)!+1=(n+1)!\cdot[n+1-1]+1=\\\\\\=&#10;n(n+1)!+1

So the answer is:

\boxed{\sum\limits_{k=0}^{n}(1+k^2)k!=n(n+1)!+1}

Sorry for my bad english, but i hope it won't be a big problem :)
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