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3 years ago

6

The volume, V, of a right cone varies jointly with the square of the radius of the base, r, and the height, h. When r = 6 and h

= 8, the volume of the cone is 96π. What is the formula for the volume of a right cone?

1 answer:

Sveta_85 [38]3 years ago

6 0

Answer:

V=πr^2(h/3)

Step-by-step explanation:

V=π6^2(8/3)

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JKL is a straight line.

kirza4 [7]

Answer:

x = 32°

Step-by-step explanation:

∆KLM is an isosceles triangle because it has two equal sides, KL & KM. Therefore, the angles opposite to each of the two equal sides, which are referred to as the base angles are congruent to each other.

m<KML = m<KLM = 58°

m<MKL = 180 - (58 + 58) (Sum of triangle)

m<MKL = 64°

m<JKM = 180 - m<MKL (linear pair theorem)

m<JKM = 180 - 64 (Substitution)

m<JKM = 116°

∆JKM is also an isosceles triangle with two equal sides. Therefore, it's based angles (x & <J) would also be equal to each other.

Thus:

x = ½(180 - m<JKM)

x = ½(180 - 116) (Substitution)

x = 32°

3 0

2 years ago

if A= (-1, -3) and B= (11, -8) what is the length of AB A) 14 units B) 11 units C) 13 units D) 12 units

Anastasy [175]

Distance is the formula :

D $AB$ = $\sqrt{(xA - xB)^2 + (yA - yB)^2}$

so

$\sqrt{(-1 - 11)^2 + (-3 + 8)^2}$

= 13

Answer is C)

5 0

2 years ago

Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=

Tom [10]

Answer:

16π

Step-by-step explanation:

Given that:

The sphere of the radius = $x^2 + y^2 +z^2 = 4^2$

$z^2 = 16 -x^2 -y^2$

$z = \sqrt{16-x^2-y^2}$

The partial derivatives of $Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}$

$Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}$

Similarly;

$Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}$

∴

$dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA$

$=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA$

$=\sqrt{ \dfrac{16}{16-x^2-y^2} }\ \ .dA$

$=\dfrac{4}{\sqrt{ 16-x^2-y^2} }\ \ .dA$

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to $\sqrt{12}$

dA = rdrdθ

∴

The surface area $S = \int \limits _R \int \ dS$

$= \int \limits _R\int \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA$

$= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \ \ rdrd \theta$

$= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr$

$= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}$

$= 8\pi ( - \sqrt{4} + \sqrt{16})$

= 8π ( -2 + 4)

= 8π(2)

= 16π

4 0

2 years ago

Evaluate. −64125−−−−√3

s344n2d4d5 [400]

Answer:

-64123.2679492

Step-by-step explanation:

I'm sorry if this is wrong!

4 0

2 years ago

Step by step Simplify: -9j(5)(3k)

SpyIntel [72]

-9j((5)(3k))
-9j(15k)
-135jk

3 0

3 years ago