Help--Math problem

[John Zuber]

Math problem: 26.54-P = .5(50-p) solve and show work for P



thanks!

 

[Michael Cacioppo]

Math problem: 26.54-P = .5(50-p) solve and show work for P



26.54-P = .5(50-p) = .25- .5p

26.54 = p - .5p +.25

26.54 - .25 = p - .5p = .5p

26.29 = .5p

p = 26.29/.5 = 52.58

 

[John Zuber]

thanks Mike

 

[Richard Kolb]

this looks like high school stuff, what is it?



Let's see here, showing work sucks on the interweb:

53.08 - 2p = 50 - p divide each side by .5 or multiply each side by 2

53.08 = 50 + p add 2p to each side

3.08 = p subtract 50 from each side



plug 3.08 in for p to check your work. Unless the p and P are different variables that's the answer.

 

[Mike Caldwell]

Alright i kinda feel like a dork but..

Here's what I got.

26.54-p=.5(50-p)

26.54-p=25-.5p (multiply .5 into brackets)

26.54-25-p=.5p (collect common terms)

1.54=p-.5p (collect common terms cont.)

1.54=.5p

1.54/.5=p

p=3.08

ok there thats what i got and i jsut checked by putting3.08 back in for p and they equal eachother. hope this helps

 

[Mike Caldwell]

Darn rich your faster than I. Yeah I think i learned it in grade 11 but im now doing it in my finance course in college so who knows.

 

[Trevor Yeomans]

2 many numbers...........:wacko:

 

[Michael Cacioppo]

I misread (50-p) as (.50-p) Sorry for the error John...MikeC





Math problem: 26.54-P = .5(50-p) solve and show work for P



26.54-P = .5(50-p) = .25- .5p

26.54 = p - .5p +25

26.54 - 25 = p - .5p = .5p

1.54= .5p

p = 1.54/.5 = 3.08

 

[Ben Poche]

2+2=4

 

[Bill V]

Actually, you're all wrong. The original problem states:



26.54-P=.5(50-p)



There are two different variables--P and p. Solving for P gives you:



-P = 25 - .5p - 26.54 (distribute the .5*(50-p), then subtract 26.54 from both sides)



P = 1.54 + .5p (Combine the numeric values, multiply both sides by -1)



THAT'S the actual answer. You can't assume that P and p are the same thing--unless you're told otherwise, they need to be considered to be two different entities.

 

[Jeff C]

Lol, very good Bill V. Man, I wish the math I do still had numbers in it.

 

[Mike Wilson]

My brain hurts just looking at that!

I guess you can tell that I am not our "Resident Mathemetician"!

Although I do well, I must say, of being the subjugator of the English Language

on this site, my math skills, in the higher context, are somewhat lacking!

And, I, do admit, that, even I, can, on occasion, over-punctualize, a seemingly,

otherwise, free-flowing, sentence!

 

[Ryan Schaecher]

Come on! Bring on the Integration and DiffEQ's!! At least I can do most of those.... forget the Algebra.

 

[Jeff C]

Here's one for ya, Shek.



An air filled rubber ball has a diameter of 6in. If the air pressure within it is increased until the ball's diameter becomes 7in, determine the average normal strain in the rubber.

 

[Bill V]

Jeff C--



Here's my very quick, not-well-thought-out attempt:



Surface area of a sphere = 4*pi*r^2 = pi*d^2



Surface area of a 6" sphere is therefore 36*pi.



Surface area of a 7" sphere is therefore 49*pi



Strain is therefore (49*pi)/(36*pi), or 1.361 (last digit repeating infinitely). Which means there is a strain of 36.1%



Am I anywhere close to right? Like I said, I did that very quickly, and could easily have a logic error somewhere. But I don't believe any information about the properties, thickness, etc., of the rubber are necessary--it's simply a geometry problem....



--Bill

 

[Ryan Schaecher]

uh... ok.... that was kind of a joke.... In my engineering classes we had the statement that give me integral, differentials, first and second order diffEq's, but just don't make me add and subtract....



I know nothing about Rubber (mathematical properties theirof). Guess I spoke too soon.... But it would seem to me that the coeffecient of the rubber, the thickness, rigidity etc would all be ncessary. What type of rubber - latex, vulcanized, etc.

 

[Bill V]

And Q--You're right, "trick" caps and non-caps problems are unheard of. But it's not because caps/non-caps problems don't exist. It's because when they exist, they're not considered to be trick questions. My college mathematics and engineering courses quite frequently used variables that differed in being only upper and lower case. Some books and professors, for example, would use R and r, instead of the R1 and R2 (with the 1 and 2 subscript) conventions that other books and professors used. Those professors/writers felt using R and r was less confusing than using variables that included numbers, as those could be misconstrued as being values, rather than part of the variables. Similary, RA and RB were inappropriate, as people could interpret them as being R*A and R*B. So they chose to just use R and r.



And after explaining all that, I need some R and r. Bill V out.

 

[Bill V]

Actually, no. As the problem is written, my answer is 100% correct--it's not a matter of opinions, it's mathematical fact.



What you're "90%" sure is wrong is the problem itself--not me or my solution.

 

[Rob T]

What is the strain on the ball with the 6" diameter? This is required to figure the strain on the ball when it is inflated to 7" diameter.

 

[TomT]

There’s three kinds of mathematicians, those that can add and those that can’t.