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juin [17]
2 years ago
15

A and b are positive integers and a-b=2 ... Evaluate the following:

Mathematics
1 answer:
lana [24]2 years ago
5 0

Answer:

Step-by-step explanation:

a-b=2

\frac{27^{\frac{1}{3} b} }{9^{\frac{1}{2}a } } \\=\frac{(3^{3}) ^{\frac{1}{3} b} }{(3^{2} )^{\frac{1}{2} a} } \\=\frac{3^b}{3^a} \\=3^{b-a} \\=3^{-(-b+a)} \\=3^{-2}\\=\frac{1}{3^2} \\=\frac{1}{9}

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Answer:

Check below, please.

Step-by-step explanation:

Hi, there!

Since we can describe eccentricity as e=\frac{c}{a}

a) Eccentricity close to 0

An ellipsis with eccentricity whose value is 0, is in fact, a degenerate one almost a circle. An ellipse whose value is close to zero is almost a degenerate circle. The closer the eccentricity comes to zero, the more rounded gets the ellipse just like a circle. (Check picture, please)

\frac{x^2}{a^2} +\frac{y^2}{b^2} =1 \:(Ellipse \:formula)\\a^2=b^2+c^2 \: (Pythagorean\: Theorem)\:a=longer \:axis.\:b=shorter \:axis)\\a^2=b^2+(0)^2 \:(c\:is \:the\: distance \: the\: Foci)\\\\a^2=b^2 \\a=b\: (the \:halves \:of \:each\:axes \:measure \:the \:same)

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5=\frac{c}{a} \:c=5a

An eccentricity equal to 5 implies that the distance between the Foci has to be five (5) times larger than the half of its longer axis! In this case, there can't be an ellipse since the eccentricity must be between 0 and 1 in other words:

If\:e=\frac{c}{a} \:then\:c>0 , and\: c>0 \:then \:1>e>0

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a=c\\\\a^2=b^2+c^2\:(In \:the\:Pythagorean\:Theorem\: we \:should\:conceive \:b=0)

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