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Parabolic Problems

By Altum Spectent
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Another problem I have seen lately is that of parabolas and symmetry, as well as that of intervals, domains, ranges, and other graphical dilemmas. If one is asked to define the interval over which a function is increasing, decreasing, or remaining the same, the answer is always (x 1 , x 2 ). One never uses brackets when referring to an interval! However, if one is asked to define a domain or range, the answer could include one or more brackets: (x 1 , x 2 ], [x 1 , x 2 ), or [x 1 , x 2 ] if referring to the domain; (y 1 , y 2 ], [y 1 , y 2 ), or [y 1 , y 2 ] if referring to the range. (x 1 , x 2 ) and (y 1 , y 2 ) can also be used to describe the domain and range, respectively, in some cases. A function is said to be increasing if the y-values are getting larger, decreasing if they are getting smaller, and constant if they are remaining steady. The domain of the function is the entire group of all the x -values for which the function is continuous, while the range is the entire
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Systems of Linear Equations

By Altum Spectent
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I'm back! Sorry for the tardiness, but college students never have much extra time on their hands. Anyway, today's post will be a mathematical review; coming directly from a recent tutoring session in which I helped a student with  systems of linear equations . Systems of linear equations can be solved by using either the substitution method, the addition method, or the graphical method. Sometimes more than one method is required, especially for systems of three equations, each with the same three variables. Personally, I prefer the addition and substitution methods, but the graphical method is fine, too. Say we are given the following system of linear equations: This is very simple and straightfor ward example. We can solve this easily using the addition method, which would simplify to 3x = 6 , which would further simplify to x = 2 . Using this result, we can solve for y : (2) - y  = 1  à  2 - 1 = y  à  y = 1 . Or 2(2) + y = 5  à  4 + y = 5  à  y = 1 . Our answer
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Chemical Mixture Problems

By Altum Spectent
Recently, I tutored a few students in math, and they were having the same problem: How to solve a chemical mixture equation? Chemical mixture problems can pretty much always be solved with the equation: p 1 (v 1 ) + p 2 (v 2 ) = p 3 (v 1  + v 2 ) , where "p" refers to the various percentages of the chemical involved and "v" refers to the various volumes of the chemical involved. The question should give either all three percentages as well as one volume, or both volumes as well as two percentages. When solving a problem such as this, it is unimportant whether the percentage is listed as a number such as 75, or a decimal, such as 0.75, as long as the same method is used throughout the problem. Here is a sample question below: A chemist needs 150 milliliters of a 50% saline solution but has only 18% and 78% solutions available. Find how many milliliters of each that should be mixed to get the desired solution. So, we can start equating variables wit
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