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3 years ago

8

Find the volume of a right circular cone that has a radius of 4 inches and a height of 12 inches. A. 36 in. B. 48 in. C. 52 in.

D. 64 in.

2 answers:

Hatshy [7]3 years ago

8 0

Answer:

D.) 64pi in.³

Step-by-step explanation:

I got it right in quiz, trust me.

I hope this helps! ^^

OLEGan [10]3 years ago

7 0

Answer:

A) 36 inches

36 Incheon is the correct answer

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How much paper will be needed to cover the cylinder shown?

Advocard [28]

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3 years ago

Read 2 more answers

Look at the triangle show on the right. The Pythagorean Theorem states that the sum of the squares of the legs of a right triang

Vladimir [108]

Answer:

$cos^2\theta + sin^2\theta = 1$

Step-by-step explanation:

Given

$(\frac{b}{r})^2 + (\frac{a}{r})^2$

Required

Use the expression to prove a trigonometry identity

The given expression is not complete until it is written as:

$(\frac{b}{r})^2 + (\frac{a}{r})^2 = (\frac{r}{r})^2$

Going by the Pythagoras theorem, we can assume the following.

a = Opposite
b = Adjacent
r = Hypothenuse

So, we have:

$Sin\theta = \frac{a}{r}$

$Cos\theta = \frac{b}{r}$

Having said that:

The expression can be further simplified as:

$(\frac{b}{r})^2 + (\frac{a}{r})^2 = 1$

Substitute values for sin and cos

$(\frac{b}{r})^2 + (\frac{a}{r})^2 = 1$ becomes

$cos^2\theta + sin^2\theta = 1$

3 0

3 years ago

Eighteen less than 7 times a number is the same as 12 more than the number

professor190 [17]

Answer:

7n-18=n+12

Step-by-step explanation:

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2 years ago

Which of the following is the equation of the one that passes through the point

Diano4ka-milaya [45]

Answer:

C

Step-by-step explanation:

the answer is C

( -6 , 1 )

y = -1/2x - 2

if x = -6

y = -1/2(-6) - 2

y = 3 - 2

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2 years ago

Suppose a white dwarf star has a diameter of approximately 1.8083 to the power of 4 km. Use the formula 4n to the power of 2 to

aivan3 [116]

ANSWER:

The surface area of the star is 3.2700 x $10^{8}$ square kilometres.

EXPLANATIONS:

Diameter of the star = 1.8083 x $10^{4}$ Km.

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Where n is the radius of the star.

So that;

n = $\frac{ 1.8083 * 10^{4} }{2}$

= 0.90415 x $10^{4}$

n = 0.90415 x $10^{4}$ Km

Thus,

Surface area = 4 x $(0.90415*10^{4} )^{2}$

= 326994889

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Therefore, the surface area of the star is 3.2700 x $10^{8}$ square kilometres.

4 0

3 years ago