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3 years ago

Given a circle with a diameter of 8, what is the circumference? a) 3p b) 4p c) 6p d) 8p

Mathematics

1 answer:

Viefleur [7K]3 years ago

4 0

I hope this helps you

Circumference = 2 .pi.r

Circumference = 2.pi.8

Circumference =16.pi

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4 years ago

write the equation in standard form for the hyperbola with vertices ( – 12,0) and (12,0), and foci ( – 13,0) and (13,0).

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Standard form for the hyperbola with vertices ( – 12,0) and (12,0), and foci ( – 13,0) and (13,0) is $\frac{x^{2} }{144}-\frac{y^{2} }{25}=1$

Hyperbola is a plane curve generated by a point so moving that the difference of the distances from two fixed points is a constant.

Given,

The Vertices of the hyperbola = (-12,0) and (12,0)

Foci = (-13,0) and (13,0)

a=12

ac=13

c= $\frac{13}{12}$

We know,

c= $\sqrt{1+\frac{b^{2} }{a^{2} } }$

$c^{2}=1+\frac{b^{2} }{a^{2} } \\(\frac{13}{12} )^{2}=1+\frac{b^{2} }{144} \\\frac{169}{144}=1+ \frac{b^{2} }{144} \\\frac{b^{2} }{144} =\frac{25}{144}\\ b^{2}=25$

The equation of the hyperbola is

$\frac{x^{2} }{a^{2} }-\frac{y^{2} }{b^{2} }=1\\ \frac{x^{2} }{144 }-\frac{y^{2} }{25 }=1$

Hence, the standard form for the hyperbola with vertices ( – 12,0) and (12,0), and foci ( – 13,0) and (13,0) is $\frac{x^{2} }{144}-\frac{y^{2} }{25}=1$

Learn more about hyperbola here

brainly.com/question/7098764

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2 years ago