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Sean K.

Math Teacher for Five Years, Math Graduate from the University of Notre Dame

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Question:

Students at Lincoln High are holding a car wash. Cars can be washed for $8 and trucks can be washed for $10 each. In the first hour, 32 vehicles were washed and $280 were made. How many trucks were washed in that first hour?

Sean K.

Answer:

In the problem above, we are given a lot of information. We know how much it costs to wash a car or a truck, we know how many vehicles there were altogether and we know how much money was made altogether. We do not know the individual amount of cars and trucks though, and the problem asks for the number of trucks specifically. Since we have two pieces of missing information, we need two variables to represent them symbolically. Let x = the number of cars Let y = the number of trucks We know that there were 32 vehicles in all, meaning the sum of the cars and trucks equaled 32. x + y = 32 But we can't figure out the number of trucks just from that information. There could be 2 trucks and 30 cars, or 15 trucks and 17 cars. We could guess the amount of trucks then check to see if that works out with the rest of the problem, but that can take quite a while. Instead, we need a system of equations to find the two missing pieces of information, x and y. Our first equation was about the number of vehicles. Since we have information on the money made, our second equation will reflect that. Each car costs $8. So two cars cost $16, three cost $24, and so on. To figure out the total amount of money made from washing cars, just multiply the number of cars washed by $8. As an expression, this looks like "8x". Similarly with the trucks, the total amount made would be "10y". Money made from both vehicles totaled $280, so we have the equation: 8x + 10y = 280 Now we have our to equations to solve. We can either use substitution or elimination. I will use elimination and multiply the first equation by -10. -10(x + y = 32) 8x + 10y = 280 -10x + -10y = -320 8x + 10y = 280 ______________ *combine the two equations, eliminating y -2x = -40 * isolate x x = 20 This means that there are 20 cars. Since there were 32 vehicles altogether, 20 cars means there are 12 trucks and we have our answer! We can even check our work, 12 trucks that cost $10 each make $120 and 20 cars at $8 each make $160. Together, $120 and $160 total $280, which matches the word problem.

Calculus

TutorMe

Question:

Differentiate y = (x^3 + 7x - 1)(5x + 2).

Sean K.

Answer:

Assuming you know basic differentiation, the challenge of this problem is that we have a product of a trinomial and a binomial. There are two major ways to differentiate the equation above. The first method is to distribute the binomial to the trinomial and then differentiate the new expression. This method is shown below. y = (x^3 + 7x - 1)(5x + 2) *rearrange to show distribution of binomial = 5x (x^3 + 7x - 1) + 2 (x^3 + 7x - 1) *distribute = 5x^4 +35x^2 - 5x + 2x^3 + 14x - 2 *combine like terms and write in descending order = 5x^4 +2x^3 + 35x^2 + 9x - 2 The derivative of the expression above is : D{5x^4 +2x^3 + 35x^2 + 9x - 2} = 20x^3 + 6x^2 + 70x + 9 Another way to differentiate the equation involves using the PRODUCT RULE. The product rule allows you to differentiate two expressions that are being multiplied together, in this case (x^3 + 7x - 1) and (5x +2). The product rule is shown with the following formula: D{ f(x)g(x) } = f(x)g'(x) + f'(x)g(x) In other words, multiply the original first factor by the derivative of the second. Then the reverse: multiply the original second factor by the derivative of the first. Add the two answers together. D { (x^3 + 7x - 1)(5x + 2) } = (x^3 + 7x - 1)D{(5x + 2)} + D{(x^3 + 7x - 1)}(5x + 2) = (x^3 + 7x - 1)(5) + (3x^2 + 7)(5x + 2) = 5x^3 + 35x -5 + 15x^3 +6x^2 + 35x + 14 = 20x^3 + 6x^2 + 70x + 9 Our answers from both methods match! As functions become more difficult to differentiate, the product rule becomes a huge help.

Algebra

TutorMe

Question:

Find the zeros (roots) of the function f(x) = (x-2)^2 -64.

Sean K.

Answer:

To find the zeros of the quadratic function listed, we first need to expand the binomial. (x-2)^2 = (x-2)(x-2) = x^2 - 4x + 4 Thus our function now becomes f(x) = (x^2 - 4x +4) - 64 If we combine the two constants (4 and -64), we obtain f(x) = x^2 - 4x - 60 Setting the function equal to zero then factoring the trinomial will allow us to find our roots. To factor the trinomial, we need to think of two factors of -60 that sum to -4. For example, two factors of -60 are 5 and -12 (5 * -12 = -60), but if we combine 5 and -12, we get -7 not -4. The correct factors are shown in the work below. 0 = x^2 - 4x - 60 0 = (x - 10)( x + 6) The two factors of our trinomial are (x - 10) and (x + 6)! Since our equation shows that the product of the two quantities is equal to zero, then one or both factors must be equal to zero as well. (Remember, whenever you multiply a number by zero, it equals zero!) This means that we should set (x - 10) equal to zero as well as (x + 6) equal to zero then solve the two equations. x - 10 = 0 (**add 10 to both sides of the equation) x = 10 x + 6 = 0 (**subtract 6 from both sides of the equation) x = -6 Thus, we have found that the zeros of our quadratic function are x = 10 and x = -6. If you have a graphing calculator and type in the original expression (x-2)^2 - 64 and take a look at the graph, you will see that the quadratic passes over the x-axis at x = -6 and x = 10. You can also substitute either value in for x in the original expression and will find that it simplifies to zero! (e.g. f(10) = (10-2)^2 - 64 = (8)^2 - 64 = 64 - 64 = 0)

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