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Find a volume of a seedless orange with radius 5cm. 523.60cm 1) suppose the peel of the orange is 0.5cm thick.What volume of the orange is eddible? 2)Express the edible part of the orange as a percent of the whole orange.Round your answer to the nearest percent. Could u explain 1) & 2) solving? Thanks.
Is the orange a perfect sphere?
lol ok, what's the volume of a sphere formula?
should look like this: v = (4/3)PI(r)^3
now for question 1) , plug in the radius of the orange for r... like this... v = (4/3)PI(5)^3
ok, so if you want to know the portion that is edible, minus the portion that is not edible, namely 0.5... so now use a radius of 4.5cm and you'll have it...
i did 5.0 - 0.5 and got 4.5
cool, so you need to know the volume of the whole orange, and the volume of the edible part to find the percentage.. Please post the two volumes, and then we'll find the percentage that is edible..
volume of the whole orange = 523.60 volume of the edible part = 380.51
volume of edible part = 381.51 i mean
523.6 is the whole and 381.7 is the part of the edible part...
how did you get 380.70?? i know that is the right answer but i keep on getting 381.51
I didn't get 380.70 I got 381.7 anyway, now we ask "what percent of 523.6 is 381.7" that equates to: x(523.6) = 381.7
wait but where is the x coming from?
now dividing both sides gives 0.72899... and moving the decimal to percent.. 72.899% and rounding... 73% of the orange is for eats.. but personally I like the peel!!! lol
oh, the x is "what percent" "what percent" is x of (means times) is means (equals) what percent of bla1 = bla2 x times bla1 = bla2 x(bla1) = bla2 x = bla2/bla1
ohhhh. hehe i personally like the of the orange. (:
You'll get it.. learn to turn what you can say into algebra expressions. It gets easy with practice just like speaking! :)