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Vaselesa [24]
4 years ago
15

What is the name of the solid figure you will make if you fold the net shown?

Mathematics
1 answer:
Sergeu [11.5K]4 years ago
4 0

Answer:hahahahh bithch4

Step-by-step explanation:

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Write an equation for an ellipse centered at the origin, which has foci at (\pm\sqrt{12},0)(± 12 ​ ,0)left parenthesis, plus min
lora16 [44]

Answer:

\mathbf{\dfrac{x^2}{49^2} +\dfrac{y^2}{37^2} =1}

Step-by-step explanation:

Given that :

the foci of the ellipse is (±√12,0) and C0-vertices are (0,±√37)

The foci are (-C,0) and (C ,0)

the focus has x-coordinates so the focus is  lie on x- axis.

The major axis also lie on x-axis

The minor axis lies on y-axis so C0-vertices are (0,±√37)

The given focus C = ae = √12

Given co-vertices ( minor axis) (0,±b) = (0,±√37)

b= √37

We can therefore express the  relation between the focus and semi major axes and semi minor axes as:

\mathbf{c^2 = a^2 - b^2 } \\ \\ \mathbf{a^2 = c^2 + b^2 } \\ \\ \mathbf{c^2 = ( \sqrt12)^2 - (\sqrt 37)^2 }  \\ \\ \mathbf{c^2 = 49 } \\ \\  \mathbf{c = \sqrt{49 }}

The equation of ellipse formula is:

\dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1

and we know that \mathbf{a=\sqrt{49}  \ \  and  \  \ b=\sqrt{37}}

Thus ; the equation of the ellipse at the origin is

\mathbf{\dfrac{x^2}{49^2} +\dfrac{y^2}{37^2} =1}

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4 years ago
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