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3 years ago

Make r the subject of the formula in 1 - E Rr

Mathematics

1 answer:

Artemon [7]3 years ago

8 0

Answer:

r= 1r- eR = r

Step-by-step explanation:

r-r = ER final answer to equal r= -ER

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Step-by-step explanation:

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MakcuM [25]

Answer:

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Step-by-step explanation:

Given equation:

$x^2+(p-3)y^2-4x+6y-16=0$

Standard equation of a circle:

$(x-a)^2+(y-b)^2=r^2$

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Therefore, $p-3=1$

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Additional information

To rewrite the given equation in the standard form:

$\implies x^2+y^2-4x+6y-16=0$

$\implies x^2-4x+y^2+6y=16$

$\implies (x-2)^2-4+(y+3)^2-9=16$

$\implies (x-2)^2+(y+3)^2=16+4+9$

$\implies (x-2)^2+(y+3)^2=29$

So this is a circle with centre (2, -3) and radius √29

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When you can divide, or multiply by the same number. for numerator and denominator.

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What is the surface area of a square pyramid with the height of 3 and base of 4

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Step-by-step explanation:

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3 years ago

Make r the subject of the formula in 1 - E Rr​

Make r the subject of the formula in 1 - E Rr