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erma4kov [3.2K]
2 years ago
11

Proving That All Circles Are Similar give: cicrcle x with radius r and circile y with radius s prove: circle x is similar to cir

cle y
Mathematics
1 answer:
sergij07 [2.7K]2 years ago
4 0

Answer:

Two shapes are similar when they have the same shape while their sizes may be the same or different

Therefore, two similar shapes are shapes that have a relationship such that the dimensions of one of the shapes can be obtained from the dimensions of the other shape by multiplying by a scale factor

All circles have the same shape and are defined by their center and radius

Given circle with center at 'x' and radius 'r' and circle with center 'y' and radius 's', then there exist a scale factor a = s/r such that we have;

r × a = s

Where a = s/r, we get;

r × s/r = s

We can therefore, obtain a circle with with the same size as the circle with center 'y' and radius 's' by multiplying the radius of the circle with center at 'x' and radius 'r' by a

Therefore the circle with center at 'x' and radius 'r' is similar to the circle with center 'y' and radius 's'

Step-by-step explanation:

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64m^3

Step-by-step explanation:

The volume of a cube formula is side^3.

Using that formula, we get 64 meters cubed as the answer.

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Select the ratios simplest form is 8/7
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Yes, a rhombus

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Hello.
melisa1 [442]

Answer:

\large{\boxed{\sf 2y^2\sqrt{2y}}}

Step-by-step explanation:

Here the given expression to us is ;

\sf\qquad\longrightarrow \dfrac{\sqrt{64xy^5}}{\sqrt{8x}}

Recall that ,

\sf\qquad\longrightarrow \dfrac{\sqrt{x}}{\sqrt{y}}=\sqrt{\dfrac{x}{y}}

On using this , we have ;

\sf\qquad\longrightarrow \sqrt{\dfrac{\cancel{64}\cancel{x}y^5}{\cancel{8x}}}

Simplify ,

\sf\qquad\longrightarrow \sqrt{ 8 y^5}

The prime factorisation of 8 is 2³ . So ;

\sf\qquad\longrightarrow \sqrt{ (2^3)(y^5)} =\sqrt{(2^2)(2)(y^2)(y^2)(y)}

Simplify the square root ,

\sf\qquad\longrightarrow \pink{ 2y^2\sqrt{2y}}

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