BloodFire101
Answered

Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Join our platform to connect with experts ready to provide precise answers to your questions in different areas. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

Can i have some help Please! (Brainliest As Well if it lets me) Pick question 16 or 17 to answer i only have to do one Be specific in your comparisons - think about K (dilation scale factor) and how K changed for circumference versus area. You will need to look up the formula for surface area of a sphere.

Can I Have Some Help Please Brainliest As Well If It Lets Me Pick Question 16 Or 17 To Answer I Only Have To Do One Be Specific In Your Comparisons Think About class=

Sagot :

facundo3141592

For problem 17, if you double the diameter of the moon, then the surface increases by a factor of 4 and the volume by a factor of 8.

How to compare the change in surface and volume?

Let's answer question 17.

Remember that for a sphere of radius R, the volume is:

[tex]V = (4/3)*pi*R^3[/tex]

And the surface area is:

[tex]S = 4*pi*R^2[/tex]

So the volume depends on the cube of the radius and the surface on the square of the radius.

Now, if we double the diameter of the moon, then we also double the radius.

So the new volume will be:

[tex]V' = (4/3)*pi*(2*R)^3 = 8*[(4/3)*pi*R^3] = 8*V[/tex]

The new volume is 8 times the original volume.

And for the surface:

[tex]S' = 4*pi*(2*R)^2 = 4*[4*pi*R^2] = 4*S[/tex]

So the new surface is 4 times the original surface.

If you want to learn more about spheres:

https://brainly.com/question/10171109

#SPJ1

Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.