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grandymaker [24]

3 years ago

15

A cone where the angle of vertex is 2pi/3. Fill in the blanks found in the attachment.

1 answer:

Firdavs [7]3 years ago

4 0

As far as I understand, you've made two mistake which do not satisfy your task. Well, I seggest you adding b = 2pi/3 to the first gap instead of what you add in the previous effort. Then, to solve the next part, I would change f on <span>8/sqrt(3)*cospi
I bet that this answer will sort everything out. Regards!</span>

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kherson [118]

Answer:

$\large\boxed{(x-2)^2+(y-1)^2=34}$

Step-by-step explanation:

The equation of a circle in standard form:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

We have the endpoints of the diameter: (-1, 6) and (5, -4).

Midpoint of diameter is a center of a circle.

The formula of a midpoint:

$\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)$

Substitute:

$h=\dfrac{-1+5}{2}=\dfrac{4}{2}=2\\\\k=\dfrac{6+(-4)}{2}=\dfrac{2}{2}=1$

The center is in (2, 1).

The radius length is equal to the distance between the center of the circle and the endpoint of the diameter.

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the coordinates of the points (2, 1) and (5, -4):

$r=\sqrt{(5-2)^2+(-4-1)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}$

Finally we have:

$(x-2)^2+(y-1)^2=(\sqrt{34})^2$

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