Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (anonymous):

A cone has a radius of 15 cm and a volume of 540 cm3. What is the volume of a similar cone with a radius of 10 cm? 54 cm3 240 cm3 160 cm3 360 cm3

OpenStudy (anonymous):

I think the answer is 240.

OpenStudy (anonymous):

please help, will give medal out to best answer.

jigglypuff314 (jigglypuff314):

If two solids are similar, then their corresponding sides are all proportional. The ratio of their surface areas is the side ratio squared and note that the ratios of the areas does not give the actual surface areas. The volume ratio for the two solids is the side length ratio raised to the third power. Again, this is not the solids' volume, only the ratio of the volumes. https://www.brightstorm.com/math/geometry/similarity/similarity-and-volume-ratios/ by that description... that means if they give you the ratio of the lengths to be a/b then the ratio of the volumes would be (a/b)^3 or a^3/b^3 so they gave you 10/15 as the lengths ratio and so what would be (10/15)^3 = ?

OpenStudy (anonymous):

that would equal 0.29.

jigglypuff314 (jigglypuff314):

I was thinking of keeping it as a fraction so we can do a ratio like (10/15)^3 = (2/3)^3 = 8/27 so ratio 8 x --- = --- solve for x 27 540

OpenStudy (anonymous):

x is 160

jigglypuff314 (jigglypuff314):

yeah... and that's what I think the answer should be... unless you were taught differently xP cuz I just taught myself the concept without a teacher so idk if I'm right ;)

OpenStudy (anonymous):

ok, thanks so much. I think I understand now how to do that from now on. could you help me with some other questions? normally I just do it myself but this test is a lot of my grade and most of the stuff i'm ok with but some i'm not very sure how to do them or the answer of them.

jigglypuff314 (jigglypuff314):

I'll just try to ignore that you told me that it's a test (since helping on tests is cheating) and I'll see if I can help xD

OpenStudy (anonymous):

k thanks so much, just give me like 2 minutes to get them ready.

OpenStudy (anonymous):

3. Which of the following are considered skew lines? http://tinypic.com/view.php?pic=k9vx8p&s=8#.U2vjG1e8hyU (1 point) AC and CD DH and BF EG and FH AE and GH

OpenStudy (anonymous):

I'll submit them one at a time if thats ok?

jigglypuff314 (jigglypuff314):

Two nonparallel lines in space that do not intersect are called Skew lines. http://www.icoachmath.com/math_dictionary/skew_lines.html

jigglypuff314 (jigglypuff314):

idk whut that means but ok xD

OpenStudy (anonymous):

http://tinypic.com/view.php?pic=fwjfvo&s=8#.U2vm21e8hyU 7. Which solid does the net form? (1 point) square pyramid triangular prism triangular pyramid cube

jigglypuff314 (jigglypuff314):

well if you think about folding it up mentally... then |dw:1399580647623:dw|

OpenStudy (anonymous):

I think d is the answer for the skew lines question and a is the answer to this one

jigglypuff314 (jigglypuff314):

correct and correct :) good job! :D

OpenStudy (anonymous):

ok thanks, now i'll submit the next one.

OpenStudy (anonymous):

http://tinypic.com/view.php?pic=2vkjmo2&s=8#.U2vpCFe8hyU 16. The cones below are similar, although not drawn to scale. What is the length of the radius of the larger cone? (1 point) 5 ft 6 ft 7 ft 8 ft

jigglypuff314 (jigglypuff314):

5 x -- = -- 15 18

OpenStudy (anonymous):

the answer is 6

OpenStudy (anonymous):

http://tinypic.com/view.php?pic=15xl8id&s=8#.U2vr8Fe8hyU 15. What is the length of the diagonal for the given rectangular prism to the nearest whole unit?

OpenStudy (anonymous):

Am i right about number 16?

jigglypuff314 (jigglypuff314):

erm... use pythagorean theorem to find diagonal produced by legs 12 and 4 then use the value that you get and 9 as the legs for another right triangle...

jigglypuff314 (jigglypuff314):

yes :)

OpenStudy (anonymous):

I get 15.52 after doing what you said which would mean the answer is 16, is that right?

jigglypuff314 (jigglypuff314):

12^2 + 14^2 = 340 340 + 9^2 = 421 square root of 421 = 20.518

OpenStudy (anonymous):

ok I understand, thanks so much for your help, there is only three more if you don't mind helping me finish.

OpenStudy (anonymous):

http://tinypic.com/view.php?pic=fbmcr8&s=8#.U2vw0le8hyU 8. What is the surface area of the given figure? (1 point) 2,520 cm2 2,792 cm2 4,080 cm2 2,280 cm2

jigglypuff314 (jigglypuff314):

find the area of each side and add it all together you have three rectangles and two right triangles area of a rectangle = length * width area of triangle = (1/2) base*height

OpenStudy (anonymous):

I'm sorry but this is one thing i am terrible at, I don't even see how to do this besides the first triangle with 24 as the base and 10 the height. Can you help me understand?

OpenStudy (anonymous):

or could you help me with this one while i try to understand the other one? 9. Use the net to find the approximate surface area of the cylinder to the nearest square meter. http://tinypic.com/view.php?pic=rjr5sx&s=8#.U2vxE1e8hyU (1 point) 440 m2 314 m2 283 m2 214 m2

jigglypuff314 (jigglypuff314):

the left rectangle is 10 by 34 the top rectangle is 26 by 34 the bottom rectangle is 24 by 34 and you're right about the triangles :)

jigglypuff314 (jigglypuff314):

the surface area of a cylinder is base circle + base circle + rectangle area of base circle = pi*radius^2 = pi*(5)^2 = ? area of rectangle = length*width = length*circumference of circle = length*2*pi*radius = 9*2*pi*5 = ?

OpenStudy (anonymous):

for number 8 the answer is d, now I'll work on the other one.

jigglypuff314 (jigglypuff314):

correct :)

OpenStudy (anonymous):

my comp just crashed i'm sorry the answer is a

jigglypuff314 (jigglypuff314):

yes :)

OpenStudy (anonymous):

Find the lateral area of the square pyramid. http://tinypic.com/view.php?pic=16gnbyp&s=8#.U2v6ole8hyU square pyramid (3 points)

jigglypuff314 (jigglypuff314):

erm sorry but idk wut lateral area is... o.o

OpenStudy (anonymous):

oh ok, that's fine, may need your help on a couple other answers cause i lost all my answers including ones that i didn't ask you.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!