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Algebra in Cell Culture

April 28th, 2016 by Steven Pomeroy | No Comments | Filed in Algebra






Cell culture is a technique used by many biological laboratories the world over.  Typically, cell culture requires a growth medium composed of fetal calf serum (FBS) and a specialized base medium.  It is not unusual to utilize basic algebra when making a batch of growth medium.  Here is a scenario where algebra can be used to solve a problem in making a batch of growth medium.

cell culture medium

Algebra is used in cell culture

Sally has only 62 milliliters (ml) of the very expensive base medium “Peabody Sassy” (yea – I made that up, so don’t bother Goggling it!) left.  Her growth medium is composed of Peabody Sassy and 10% FBS.  If she uses all 62 ml of Peabody Sassy (she does not want to waste any, after all), how many ml of FBS will she need to make up her growth medium?

Please enter your solutions in the message box below, and be sure to show how you arrived at it.  Sure, you can get a good estimate by trial and error, but you can get an exact answer (and faster!) by using algebra.

 

 

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3rd Grade Homework – The Distributive Property

February 21st, 2016 by Steven Pomeroy | No Comments | Filed in 3rd Grade Homework

My son came home from school recently with some 3rd grade math homework that he was having trouble with.  The homework assignment involved the Distributive Property, which for many of us has been long forgotten about.  Anyways, the homework assignment perked my interest because it showed students how the Distributive Property can be used to make difficult multiplication problems very easy.

 

First, to refresh your memory, the Distributive Property is an algebraic property that describes how a multiplication problem such as A(B + C) can be solved.   Simply, the number outside of the parentheses must be distributed across the numbers within the parentheses:

 

A(B + C) = AB + AC

 

Does this look familiar?  I hope so.  So, how does this help us to solve tough multiplication problems?  Let me give you an example:

 

8(3 + 5)

 

Of course you can just add 3 to 5 and get the answer 64 very quickly if you know your multiplication tables.  Now, by using the Distributive Property:

 

8(3 + 5) = 24 + 40 = 64

 

So, we got the same (correct) answer using the Distributive Property.  So how can we use this property to solve difficult multiplication problem?  Well, if you use the method illustrated above in reverse, you can!  Here is an example:

 

8 x 13

 

Well, not very difficult, but a good example on how to apply the technique.  First, we can break down the “13” portion of the problem to “10 + 3”:

 

8 x (10 + 3)

 

By the Distributive Property:

 

8 x (10 + 3) = (8 x 10) + (8 x 3)

 

So the problem now becomes WAY more easier than 8 x 13:

 

(8 x 10) + (8 x 3) = 80 + 24 = 104

 

So, how about a herder problem?  Let’s try 7 x 1066.  Here, we will break up 1066 into more than two parts:

 

7 x 1066 = 7(1000 + 10 + 50 + 5 + 1) = (7000 +70 + 350 + 35 + 7)

 

This is a rather simple addition problem that you can solve with pen and paper, or even do in your head, with the answer being 7462.  Cool!!

 

How about trying these on your own:

 

 

4 x 528

 

9 x 156

 

8 x 252

 

11 x 1287

 

 

 

 

 

 

 

 

 

 

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Bachet Weighing Problem

June 3rd, 2015 by Steven Pomeroy | No Comments | Filed in Historical Math Puzzles



This puzzle goes weigh way back, several hundred years, and was devised by Claude Gaspard Bachet de Méziriac.

Claude Gaspard Bachet de Méziriac

Claude Gaspard Bachet de Méziriac

His puzzle does take some thought, and you have to be careful with the wording:

What is the least number of weights that can be used on a scale pan to weigh any integral number of pounds from 1 to 40 inclusive, if the weights can be placed in either of the scale pans?

Good luck!