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# small cubes with edge lengths of 1/4 will be..

## Tags: Mathematik

18:26 Uhr, 17.05.2022

Tag!
Small cubes with edge lenghts of $\frac{1}{4}$ will be packed into the right rectangle prosm. how many small cubes are needed to fill the right rectangular prism

the formula for a right rectangular prism is
$A\phantom{\rule{0.12em}{0ex}}=2$ (wl+hl+hw)
can $i\phantom{\rule{0.12em}{0ex}}$ use this formula

$l\phantom{\rule{0.12em}{0ex}}=4\frac{1}{2}$ in
$w\phantom{\rule{0.12em}{0ex}}=5$ in

$h\phantom{\rule{0.12em}{0ex}}=3\frac{1}{4}$ in

and then plug in the dimensions given and solve for the area?
then see how many cubs of $\frac{1}{4}$ in are needed to fill the area?

18:31 Uhr, 17.05.2022

Hello,

is there a picture to this exercise? If yes, please attach it.
As far as I see you have to fill the rectangular prism with the cubes. So you need the formula for the volume.

greetings,
pivot

18:45 Uhr, 17.05.2022

Oh, it is volume then. okay,
$V\phantom{\rule{0.12em}{0ex}}=l\phantom{\rule{0.12em}{0ex}}\cdot w\phantom{\rule{0.12em}{0ex}}\cdot h\phantom{\rule{0.12em}{0ex}}$
$V\phantom{\rule{0.12em}{0ex}}=4\frac{1}{2}\cdot 5\in \cdot 3\frac{1}{4}$

$\frac{9}{2}\cdot 5\cdot \frac{13}{4}$

Volume $=73.13{\in }^{3}$

so now, how doi calculate how many cubas fit in this space?

can $i\phantom{\rule{0.12em}{0ex}}$ perform division, as in,
$73.\frac{13}{1}/4$

is this correct?

18:57 Uhr, 17.05.2022

You approach is too complicated. The length of a cube is $\frac{1}{4}.$

How many cubes fits in a row with a height of $3\frac{1}{4}=\frac{13}{4}$?
How many cubes fits in a row with a length of $5=\frac{20}{4}$?
How many cubes fits in a row with a width of $4\frac{1}{2}=\frac{18}{4}$?

Then use that the volume is $V\phantom{\rule{0.12em}{0ex}}=h\phantom{\rule{0.12em}{0ex}}\cdot l\phantom{\rule{0.12em}{0ex}}\cdot h\phantom{\rule{0.12em}{0ex}}$

>>9/2⋅5⋅13/4

Volume =73.13∈3<<

Here it is better to keep the fractions:
$9/2\cdot 5\cdot 13/4=18/4\cdot 20/4\cdot 13/4$

And you can go on to replace the numbers by the corresponding number of cubes.

19:05 Uhr, 17.05.2022

you wrtoe

$4\frac{1}{2}=\frac{18}{4}$ ??????

you meant $\frac{9}{2},$ nicht war?

19:09 Uhr, 17.05.2022

I've made an edit which refers to your approach. Have a look.

Isn't 18/4=9/2?

19:09 Uhr, 17.05.2022

and pivot, please, you said my approach was too complicated. but that does not mean it is wrong, or is it?

if it is not?
then, can I divide $73.13$ by $\frac{1}{4}$ to get the amount of cubes needed, please, confirm this or deny it, please, before I act on your proposed way.

19:10 Uhr, 17.05.2022

I've made an edit. Have a look. (18:57)

19:47 Uhr, 17.05.2022

pivot, I like your approach, this one
how many $\frac{1}{4}$ in cubes fit in a row with a height of $3\frac{1}{4}$?

what operation Must I do to solve this?

is it division, like in $\frac{13}{4}$ ÷ $\frac{1}{4}$?

20:04 Uhr, 17.05.2022

Yes. And $13/4÷1/4=13/4*4/1=\frac{13}{4}*4$

I post a slightly other way, although we are not finished yet with this.

The volume of the small cube with the side length 1/4 is $\frac{1}{{4}^{3}}$

The volume of the big cube is $18/4\cdot 20/4\cdot 13/4=\frac{18\cdot 20\cdot 13}{{4}^{3}}$

Therefore $18\cdot 20\cdot 13$ small cubes fit in the big cube.

20:20 Uhr, 17.05.2022

$\frac{13}{4}$ ÷ $\frac{1}{4}=\frac{13}{4}\cdot \frac{4}{1}=\frac{52}{4}=13$

$\frac{1}{4}$ ÷ $5=\frac{1}{4}\cdot \frac{5}{1}=\frac{5}{4}=1.2$

$\frac{9}{2}$ ÷ $\frac{1}{4}=\frac{9}{2}\cdot \frac{4}{1}=\frac{36}{2}=18$

so I will add all the results

$13+1.2+18=32.20$

is this correct, if so, then what is next?

20:34 Uhr, 17.05.2022

I am missiong something. $i\phantom{\rule{0.12em}{0ex}}$ am sorry. ich verstehe nicht

20:38 Uhr, 17.05.2022

you told me miy way was too complicated then you psoted how many $\frac{1}{4}$ in cubes fit in a row with a height of $3\frac{1}{4}=$ we agreed it was $\frac{13}{4}$

and we did this for the three dimensions

$5\cdot \frac{4}{1}=20$

$\frac{9}{2}$ ÷ $\frac{1}{4}=\frac{9}{2}\cdot \frac{4}{1}=\frac{36}{3}=12$

so we have $\frac{13}{4},20,12,$

what do we do with these?. I lost you here.

20:45 Uhr, 17.05.2022

isn't this way even better?

lets find the volume of the small cube

$V\phantom{\rule{0.12em}{0ex}}=l\phantom{\rule{0.12em}{0ex}}\cdot w\phantom{\rule{0.12em}{0ex}}\cdot h\phantom{\rule{0.12em}{0ex}}$

$=\frac{1}{4}\cdot \frac{1}{4}\cdot \frac{1}{4}$

$\frac{1}{64}$
volume of the smaoll cube $\frac{1}{64}$

let's find the volume of the bigger rectangular prism

$V\phantom{\rule{0.12em}{0ex}}=l\phantom{\rule{0.12em}{0ex}}\cdot w\phantom{\rule{0.12em}{0ex}}\cdot h\phantom{\rule{0.12em}{0ex}}$
$V\phantom{\rule{0.12em}{0ex}}=\frac{4\cdot 1}{2}\cdot 5\cdot \frac{3\cdot 3}{4}=\frac{675}{8}$

now,
$\frac{675}{8}\cdot \frac{64}{1}=5400$
$5,400$ cubes of $\frac{1}{4}$ in will ft in to the rectangular prism.

Isn't this a better way?

20:48 Uhr, 17.05.2022

Second line:

You have to divide it always by 1/4 and not the other way round. Thus we have 5&divide;1/4=5*4=20.

The rest is right.

20:56 Uhr, 17.05.2022

You have to divide it always by $\frac{1}{4}$ and not the other way round. Thus we have 5&divide;1/4=5*4=20.

but pivot doesn't division turns into multiplication ????

as in,

5 ÷ $\frac{1}{4}$ becomes $5\cdot \frac{4}{1}$ which in the end is the same thing $=20$

21:08 Uhr, 17.05.2022

Xes, you're right.

21:25 Uhr, 17.05.2022

But it is not the same as $¼÷5=1/20.$

22:03 Uhr, 17.05.2022

Alright. Ich jetzt verstehe. thanks, pivot!.