Blood Algebra

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Those who enter these career paths must know and use algebra to produce and price items, assure that buildings are constructed safely and according to plans and properly administer drugs and fitness programs. Careers in the business field include banking, financial planning, business operations and accounting. Algebraic equations using time and interest rates as variables help bankers, lenders and financial planners calculate loan payments, the value of savings accounts and certificates of deposits at a future point in time and the gain or decline on investments.

Business owners and financial officers use algebra to set prices according to desired profit margin and the cost of producing or acquiring an item or service for sale. Accountants depend upon a basic accounting equation -- stating that assets equal liabilities plus equity -- to track transactions and their effect on a business. Fitness trainers and instructors show clients proper exercise techniques, design exercise programs, provide information and set goals for weight loss and other indicators of fitness. As noted by the American Council for Exercise, algebraic formulas allow fitness professionals to figure a person's current body fat and what amount of weight loss will achieve a desired body-fat percentage.

Fitness trainers may use algebra to determine the ratio of diet to exercise needed for achieving and maintaining weight loss. Civil engineers design, inspect and maintain roads, bridges, tunnels, water supply and sewage-treatment systems, buildings and other parts of major construction projects. These professionals rely on algebra to calculate how much weight roads and structures can hold and their capacity to withstand high winds, earthquakes, snow and other elements.

For example, the Contra Costa County Office of Education illustrates how an engineer can determine the width of a bridge by using an equation that states that maximum capacity equals the bridge's length times its width times the distribution of weight. Architects design buildings and draw building plans. Their blueprints depict the length, height and width of rooms, doors, windows, floors and other components of a building to a scale.

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In a scaled drawing, an inch or fraction thereof represents a particular number of feet or yards. Architects need algebra to convert the inch or fraction to feet or yards so that the drawing has the correct proportions. Medical professionals need skills in algebra to fill prescriptions, administer drugs and detect potential heart problems and other illnesses. Heart doctors measure cardiac output, or the volume of blood the heart can pump, by using algebraic equations based on the oxygen levels in the body and heart rate.

Physicians, nurses and pharmacists must correctly decide the concentrations of medicines to administer to patients. Algebraic equations assist physicians in knowing when a prescription loses its effectiveness and should be discarded. According to Ross University, medical professionals rely on algebra to convert units of measurement and ratios. Algebra assists chefs in having the right amount of ingredients to prepare food and serve diners. As Rice University's "Linear Algebra" paper shows, chefs can use a matrix to identify foods on a menu, the ingredients and the quantity of each ingredient and then multiply the matrix by the number of diners.

Restaurant owners use algebra to calculate the cost of cooking a dish and determine how to price the dish. So let's see if we can somehow take this information and apply it to this equation. So this tells us that after one half-life-- so t is equal to 5, N of 5, is equal to the amount we start off with.


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So we're starting off with, well, we're starting off with N sub 0 times e to the minus-- wherever you see the t you put the minus 5, so minus k, times 5, That's how many years have gone by. So if we try to solve this equation for k, what do we get?

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Divide both sides by N naught. If we take the natural log of both sides, what do we get? The natural log of e to anything, the natural log of e to the a is just a. I just took the natural log of both sides. The natural log and natural log of both sides of that. But let's see if we can do that again here, to avoid-- for those who might have skipped it. So it equals 1.

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So now we have the general formula for carbon, given its half-life. At any given point in time, after our starting point-- so this is for, let's call this for carbon, for c the amount of carbon we're going to have left is going to be the amount that we started with times e to the minus k. This is our formula for carbon, for carbon If we were doing this for some other element, we would use that element's half-life to figure out how much we're going to have at any given period of time to figure out the k value. So let's use this to solve a problem.

Let's say that I start off with, I don't know, say I start off with grams of carbon, carbon And I want to know, how much do I have after, I don't know, after years? How much do I have? Well I just plug into the formula. N of is equal to the amount that I started off with, grams, times e to the minus 1. So what is that? So I already have that 1. So let me say, times equals-- and of course, this throws a negative out there, so let me put the negative number out there.

So there's a negative.

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And I have to raise e to this power. So it's 0. So this is equal to N of The amount of the substance I can expect after years is equal to times e to the minus 0. And let's see, my calculator doesn't have an e to the power, so Let me just take e. I need to get a better calculator.

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I should get my scientific calculator back. But e is, let's say 2. So this is equal to grams. So just like that, using this exponential decay formula, I was able to figure out how much of the carbon I have after kind of an unusual period of time, a non-half-life period of time. Let's do another one like this. Let's go the other way around.

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Let's say, I'm trying to figure out. Let's say I start off with grams of c And I want to know how long-- so I want to know a certain amount of time-- does it take for me to get to grams of c? So, you just say that grams is how much I'm ending up with. It's equal to the amount that I started off with, grams, times e to the minus k. That's minus 1. And now we solve for time.


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