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3 years ago

I need help please with this math problem

Mathematics

1 answer:

Arada [10]3 years ago

8 0

When the dilation is centered on the origin, you can multiply each of the individual coordinates by the scale factor.

All the coordinates are ±5 and the scale factor is 1/5. Multiplying those gives ±1. The only selection with all coordinates being ±1 is
.. Selection A.

You can check to see if the signs agree in detail. (They do.)

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Read 2 more answers

The circumference of the ellipse approximate. Which equation is the result of solving the formula of the circumference for b?

Serhud [2]

Answer:

$b = \sqrt{\frac{C^{2} }{2(\pi )^{2} } - a^{2}}$

Step-by-step explanation:

Given - The circumference of the ellipse approximated by $C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }$ where 2a and 2b are the lengths of 2 the axes of the ellipse.

To find - Which equation is the result of solving the formula of the circumference for b ?

Solution -

$C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }\\\frac{C}{2\pi } = \sqrt{\frac{a^{2} + b^{2} }{2} }$

Squaring Both sides, we get

$[\frac{C}{2\pi }]^{2} = [\sqrt{\frac{a^{2} + b^{2} }{2} }]^{2} \\\frac{C^{2} }{(2\pi)^{2} } = {\frac{a^{2} + b^{2} }{2} }\\2\frac{C^{2} }{4(\pi)^{2} } = {{a^{2} + b^{2} }$

$\frac{C^{2} }{2(\pi )^{2} } = a^{2} + b^{2} \\\frac{C^{2} }{2(\pi )^{2} } - a^{2} = b^{2} \\\sqrt{\frac{C^{2} }{2(\pi )^{2} } - a^{2}} = b$

∴ we get

$b = \sqrt{\frac{C^{2} }{2(\pi )^{2} } - a^{2}}$

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3 years ago