Solve V = π r2h for h. Show all work.
Help?

- anonymous

Solve V = π r2h for h. Show all work.
Help?

- schrodinger

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- anonymous

welcome to openstudy

- anonymous

r = 1.35
h = 12.7

- dan815

hi

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## More answers

- dan815

why dont you substitute those numbers into the formula and solve it

- tkhunny

Divide by \(\pi r^{2}\). \(\left(\pi r^{2}\right)\) is attached to 'h' via multiplication. Detach it via division.

- anonymous

@dan815 i clearly have done so already, but not coming up with the 'correct' answer according to the rest of the problem i'm inserting it into, hence why I'm seeing if im doing it wrong.

- anonymous

@tkhunny hi ther, thanks, but i'm not trying to solvefor h, i'm subsituting the numbers i posted above. . . seem to be coming up incorrect according to the book/problem set

- dan815

what is the object you are trying to find the integral of?

- dan815

volume of* sorry

- dan815

is it a cylinder?

- tkhunny

?? "Solve V = π r2h for h"
Perhaps we're speaking a different language? One might ask, why are nyou not trying to solve for h?
Are you studying Significant Figures"? What answer are you giving that is being rejected?

- bradely

step by step answers posted here
http://www.mathskey.com/question2answer/2063/solve-v-%CF%80-r-2-h-for-h
ask your Algebra1 homework questions athttp://www.mathskey.com/question2answer/algebra-1 and get free math help. all the best

- anonymous

its a beaker filled with a certain amount of liquid, all a part of a stream of different question s built off this one

- anonymous

@tkhunny because i HAVE h already, as posted above, im trying to find V

- anonymous

this is the complete problem : 5. Calculate the volume of the 50mL graduated cylinder using your measurements of diameter and height, using the formula V = ∏ r2h (r=½ diameter). This is your experimental value. Assuming the accepted value of the volume of the graduated cylinder is 50.00 mL, calculate the percent error of your volume calculation, using the following formula:
Percent Error = /accepted value - experimental value/x 100
accepted value

- dan815

lets compare answers i get for
r = 1.35
h = 12.7
pi*2*r*h = 107.725212092

- tkhunny

That's okay, but I'm really curious how this is consistent with "Solve V = π r2h for h".
Anyway, it appears we ARE studying Significant Figures. Now answer the question. What value are you using that is being rejected?

- dan815

that is area tho i think you mean pi * r^2 *h = 72.7145181618

- anonymous

i'm getting 72.7145. . .but when i plug it into the next step in my problem it doesnt make any sense! im getting a negative percentage error

- dan815

it says its pir2h so u shud use the other one 107.725212092

- dan815

your question is wrong i think
it shud say pir^2*2 not pi*r2h

- tkhunny

Well, a negative value isn't possible if you are using the absolute values. Are you using the absolute values?
Also, 72.71415 is no good. Use 72.7

- dan815

pi*r^2 *h
|dw:1362879713114:dw|

- anonymous

im totally stumped this is the exact problem taken off the homework, nothing seems to fit

- dan815

i will help you okay!! if you let me help you

- dan815

what is your R and H?

- anonymous

@dan815 ok thank you. . .im given D = 2.7 and H = 12.7 and r is 1/2 of D

- tkhunny

Well, 72.7 mL is a rather horrible approximation for a 50.0 mL beaker. Maybe something else went wrong. Or, maybe we're just testing the %-Errors formula.
\(\dfrac{|50.0 - 72.7|}{50.00}\cdot 100\;=\;\dfrac{|- 22.7|}{50.00}\cdot 100\;=\;\)

- dan815

ok your V = 1.35^2pi*12.7 = 72.7
(72.7-50/50)*100 = ansswer to 2 decimal places

- dan815

45.43%

- dan815

u understand?

- anonymous

its a negative number because its the accepted (50) minus the experiemental number, not the other way around
thats why im not understanding

- dan815

wuts a negative number?

- dan815

is the percentage negative?

- anonymous

the result is negative

- dan815

oh okay then its -45.43%

- anonymous

how is it possible to have a regative as an answer to this particular question tho?

- dan815

its possible
if its -'ve then it means ur value was 45.43% more than the actual value if its positive then its 45.43% less than the actual value

- dan815

though i think your question should say do the absolute value of the percentage of error

- dan815

but if they are defining the percentage of error without absolute value then u can get negative values

- tkhunny

You are overlooking the Absolute Values.
|50.0 - 72.7| = |72.7 - 50.0| = 22.7
There cannot be a negative value resulting from this operation.

- anonymous

ooohh ok i have no idea i havent been taught absolute values, just thrown this problem set today

- dan815

those 2 lines around the numbers mean take the absolute value so whatever the value they are always take the positive of it

- anonymous

i was given this: Percent Error = /accepted value - experimental value/x 100
accepted value
written exactly as it is, no boxes or lines around etc

- dan815

the equation for absolute value is this
|dw:1362882330695:dw|

- dan815

always gives positive answer

- dan815

those / / mean absolute value

- tkhunny

What are those slanty lines? Those are supposed to be Absolute Value Symbols. Trust me on this.

- dan815

i wish u were asian right now-_-

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