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[river shard]

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[shut musk]

so to set it up, we know the volume of the cylinder is ${pi}r^2h$

[steep bison]

;(

so to set it up, we know the volume of the cylinder is $\{pi}r^2h$
```Compilation error:```! Extra }, or forgotten $.
l.49 ... know the volume of the cylinder is $\{pi}
                                                  r^2h$
I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.

Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2023/texmf-var/fonts/map/pdftex/updmap/pdftex.map}{/usr/l```

[wispy shard]

?

[shut musk]

it says it is drilled through

so there is a cylinder in the prism

[wispy shard]

There's no $

[shut musk]

no that is for latex lol

dw

[plush bobcat]

Find the area of the hexagon, then a hexagonal "prism" or "cylinder" has a volume = area*length

[shut musk]

wrong question

[wispy shard]

Uhh

[shut musk]

with q1 part i

[wispy shard]

U can answer if u sanna

Wanna

[shut musk]

but anyway

the inner cylinder has a volume of

[plush bobcat]

For future questions I HIGHLY recommend using the crop tool that you have in your phone if it's a model made in the last 10 years

[shut musk]

$85(\pi)r^2$

[steep bison]

;(

[wispy shard]

Notedd

[shut musk]

where 85 is the height of the cylinder

[wispy shard]

Is it the prism??

[shut musk]

no

the volume of the cylinder on the inside of the prism

the “solid removed”

volume of remaining solid is

[wispy shard]

Why is 85 in front tho

[shut musk]

volume of prism minus volume of cylinder

[shut musk]

it’s just aesthetically pleasing, doesn’t really matter where you put it

[wispy shard]

Uhh

I kinda still don't get it

85 is the height and pir^2 is the volume

[shut musk]

$\frac{62^2}{2}*85-85(\pi)r^2$

[steep bison]

;(

[shut musk]

no

[wispy shard]

Oh

[shut musk]

volume is (pi)r^2*h

h is the height

[wispy shard]

Ohhh

[shut musk]

see?

[wispy shard]

Yes

[shut musk]

okay

[shut musk]

so we established this

[wispy shard]

Yes

[shut musk]

now we know the ratio is 1:6

so we get

[wispy shard]

Yes

[shut musk]

$85(\pi)r^2=6(163370-85(\pi)r^2)$

[steep bison]

;(

[shut musk]

does this make sense

[wispy shard]

Yes

[shut musk]

okay so now comes the algebra

[wispy shard]

The equal sign is the ratio??

[shut musk]

technically yes

i’m double checking whether it’s correct or not and it is

anyways

[wispy shard]

Sure

[shut musk]

$7(85(\pi)r^2)=6(163370)$

[steep bison]

;(

[shut musk]

i just uses distributive law

[wispy shard]

Why 7

[shut musk]

because on the rhs

it becomes 6(85pir^2)

and when i moved this to the other side

i added the 1 already existing on the lhs

and 1+6=7

[wispy shard]

Should it be negative

Ohh

[shut musk]

no because i added, not subtracted

[wispy shard]

Ohhh

[shut musk]

makes sense right?

[wispy shard]

Yes

Wait

[shut musk]

so you can solve for r from here right?

what

[wispy shard]

It's 6 bc it doesn't move right

[shut musk]

??

on the lhs?

[wispy shard]

Yes

[shut musk]

no

[wispy shard]

Ohhhhhhhhhhh

Wait

Nvm

[shut musk]

there was already a 85(pir^2) term existing before i moved the term from the rhs

yes

[wispy shard]

I get it

[shut musk]

now solve for r using this

use a calculator

[wispy shard]

Do we round off to 5s.f

[shut musk]

wtv u want to do

use s.f tho

if you think it’s best

so yes round 5 s.f

[wispy shard]

Ohh

Thank u so much

[shut musk]

yw

thank me in #helper-nominations and you can close the forum by using +close

if you need any more clear-ups you can dm me

[shut musk]

+close