Mathematics College

A right cone has a slant height of 6 and a radius of 4. What is its surface
area?
A. 40 units^2
B. 56 units^2
C. 24 units^2
D. 16 units^2

A Right Cone Has A Slant Height Of 6 And A Radius Of 4. What Is Its Surfacearea?A. 40 Units^2B. 56 Units^2C.

Answers

Answer 1
The answer is A! Hope you have a nice day <3
Answer 2

The surface area of the cone is 16π square units.

What is the surface area of the cone?

The surface area of the cone is given by;

⇒ Area = πrl + πr²

Here, We have;

A right cone has a slant height of 6 and a radius of 2.

Substitute all the values in the formula

⇒ Area = πrl + πr²

⇒ Area = π x 2 x 6 + π x 2²

⇒ Area = 16π

Hence, The surface area of the cone is 16π square units.

Learn more about the surface area here;

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Related Questions

A boat travels 176 miles in 13
hours (with a constant speed).
How far can it travel in 22 hours
(with the same speed)?

Answers

Answer:

297.84 miles

Step-by-step explanation:

176 miles in 13 hours is 13.5 miles per hour

13.5 * 22 = 297.84

This is the whole picture can someone help??

Answers

Step-by-step explanation:

I can't see good the picture


please i will mark you brill help me

Answers

Answer:

2n + 20

Step-by-step explanation:

2n + 20

2 times anything doubles it, then you are adding 20. An expression doesn’t have an equal sign (equations do)

Pleaseee helpp don’t understand!!

Answers

Answer:

Step-by-step explanation: so

First you should have paired attention and I don’t understand either sorry!

A slope field produces...

A. the graph of the general solution to the differential equation

B. the graph of the derivatives of the differential equation

Please explain. I am so confused on this.

Answers

It's A. B doesn't really make sense ("derivative of the differential equation" is somewhat nonsensical, "derivative of an equation" is not meaningful).

More to the point: Slope fields are used to visualize solutions to differential equations of the form

y' = f(x, y)

You take some point (x, y) and evaluate y' at the point. This gives the slope of the line tangent to the particular solution to the DE that passes through the point (x, y ).

Sample several points and evaluate y' at those points and you get several different slopes.

Simple example:

y' = x ² - y ² = (x + y ) (x - y )

Let's take the points (1, 1), (-1, 0), and (2, -2), at which we get slopes

y' = f (1, 1) = 0

y' = f (-1, 0) = 1

y' = f (2, -2) = 0

From here, you can get particular solutions that pass through a certain point by interpolating the slopes of the tangents to the solution. I've attached a slope field for the example here at the points listed above. (In the order red, green, black). Each light gray arrow in the background shows the slope of the tangent line.

If you're still unsure how the slope field is generated, I suggest looking up videos on the subject. The process is a bit difficult to describe without a dynamic visual aid.

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