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Mathematics 8 Online
OpenStudy (anonymous):

if i have a 165.1cm length and 2cm diameter of bamboo, and i want to produce 22cm length, .5cm height, and 2 cm width stick. How many possible output or stick, i will produced in my required L,W and H?

OpenStudy (thecatman):

what grade are you in?

OpenStudy (anonymous):

6 sir

OpenStudy (anonymous):

do yo any idea, can u help me @TheCatMan

OpenStudy (thecatman):

is that worded exactly as it is on the work page

OpenStudy (anonymous):

yup,

OpenStudy (anonymous):

check my work and correct me if im wrong,

OpenStudy (anonymous):

for the length, \[\frac{ 165.2 cm }{ 22cm } = 8 pcs of length\]

OpenStudy (anonymous):

for the width, i will get the circumference of the bamboo, C=piD

OpenStudy (anonymous):

C= 2 pi , then 6.28 will be the answered but im considered this as 7 cm then\[\frac{ 7 cm}{ 2cm }= 4pcs\]

OpenStudy (anonymous):

likewise, for the thickness or height, \[\frac{ 2cm }{ .5cm} =4 pcs\]

OpenStudy (anonymous):

then, do you have any idea for my next step ?

OpenStudy (thecatman):

sorry my pc froze and you have the right set up

OpenStudy (thecatman):

and i need coffee brb

OpenStudy (anonymous):

i dont know what my net step dou u have any idea for the possible output

OpenStudy (thecatman):

just put that if it doesn't require simplification

OpenStudy (anonymous):

@mathmate

OpenStudy (anonymous):

is there's any way to solve for the possible output? @TheCatMan

OpenStudy (thecatman):

im not the best to ask im not good at math my main thing is science

OpenStudy (anonymous):

ok sir thank you sir for your time :)

OpenStudy (thecatman):

im 16 please dont call me sir it makes me feel old

OpenStudy (mathmate):

@melmel

OpenStudy (mathmate):

|dw:1442797622006:dw| Yes, I think the 0.5 cm is the thinkness of the bamboo (about 1/4 of its diameter). So as you did, find the number of pieces \(N_L\) over the length by dividing 165.1cm by 22cm (reject remainder of division). Then number of pieces \(N_C\) over the circumference (6.28 divided by 2 cm). The total number of sticks will then be \(N_L*N_C\).

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