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Pani-rosa [81]
3 years ago
12

What is the value of x?

Mathematics
1 answer:
Vladimir [108]3 years ago
5 0
We know that

angle F=180°-(40+87)°=53°
applying the law of sines
EF/sin G=EG/sin F
EF=x
EG=13 cm
G=87°
F=53°
so

x/sin 87°=13/sin 53°---------> x=sin 87*(13/sin 53°)-------> x=16.3 cm

The answer is
x=16.3 cm
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Step-by-step explanation:

x/8+3=9

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The vertex angle of an isosceles triangle measures 40°. What is the measure of a base angle?
disa [49]
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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
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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)

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

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

Then:\\\\a=c\\e=\frac{c}{a}\therefore e=1

7 0
3 years ago
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krok68 [10]

Answer:

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I hope this helps you



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