![lillovecandy](/avatars/37633.jpg)
lillovecandy
29.06.2020 •
Mathematics
A right triangle ABC with sides 5 cm , 12 cm and 13 cm is revolved about the side 12 cm . Find the volume of solid so obtained.
Solved
Show answers
More tips
- W Work and Career 10 Best Ways To Find A Job: Tips To Land Your Dream Job...
- L Leisure and Entertainment How to Choose a Program for Cutting Music?...
- A Auto and Moto How to choose the right drive for your BMW...
- L Leisure and Entertainment How to Choose the Perfect Gift for Men on February 23rd?...
- H Health and Medicine How to Treat Whooping Cough in Children?...
- H Health and Medicine Simple Ways to Lower Cholesterol in the Blood: Tips and Tricks...
- O Other How to Choose the Best Answer to Your Question on The Grand Question ?...
- L Leisure and Entertainment History of International Women s Day: When Did the Celebration of March 8th Begin?...
- S Style and Beauty Intimate Haircut: The Reasons, Popularity, and Risks...
- A Art and Culture When Will Eurovision 2011 Take Place?...
Answers on questions: Mathematics
- M Mathematics Give a brief explanation why the adjusted balance method produces the least interest per cycle and why the previous balance method produces the greatest....
- M Mathematics Answer the following question images Answer like this If the question number on the picture is 9 and the answer was b 9b answer like this please...
- M Mathematics Evaluate the function f (x) = 1/2 x + 3 for f(4). a. 5 b. 9 c. 11 d. 14...
- C Computers and Technology Write a class called FastQueue to implement a queue of Strings -- therefore, the underlying array needs to be a String[] queue; Your method should also include:...
- M Mathematics What is 5.5 inches as a mixed fraction...
- C Computers and Technology PLEASE HELP I NEED HELP QUICK...
Ответ:
753.6 cm³
Step-by-step explanation:
Formed is the cone with height of 5 cm and radius of 12 cm
r= 12 cm
h= 5 cm
V=1/3πr²h= 1/3*3.14*12²*5= 753.6 cm³
Ответ:
If there are any answer choices, please state them so that people may answer.