Properties of Parallelograms<br><br> find x.

zheka24 [161]

hope you can understand

5 0

2 years ago

How would I find the range btw im only allowed a Ti-84 calculator So pls tell me how to solve this without using any stepbystep

jek_recluse [69]

It would be c because when you multiply each number by r you can see the number

7 0

2 years ago

The eccentricity e of an ellipse is defined as the number c/a, where a is the distance of a vertex from the center and c is the

Anna71 [15]

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)

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

b) Eccentricity =5

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

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:

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

c) Eccentricity close to 1

In this case, the eccentricity close or equal to 1 We must conceive an ellipse whose measure for the half of the longer axis a and the distance between the Foci 'c' they both have the same size.

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