Aliyah had $24 to spend on seven pencils. After buying them she had $10. How much did each pencil cost?
She has $24 to spend = Bought 7 pencils of unknown cost + leaving her with $10 $24 = 7p + $10 Now that you have your equation, solve for p (the price of each pencil). 24 = 7p + 10 ; Subtract 10 in each side. 14 = 7p ; Undo the multiplication with division (Divide 7 on both sides.) 2 = p Each pencil costs $2!
What is the 50th derivative of cos x?
Now, we are not just going to sit here and take 50 derivatives of cos x. That's ridiculous. If this is on your test, you wouldn't take 50 derivatives of cos x even if you wanted to because you wouldn't have the time to take 50 derivative of cos x. Instead, we're going to do what I love to do whenever solving problems in mathematics: recognize a pattern and take a shortcut. 1st derivative: -sin x 2nd derivative: -cos x 3rd derivative: sin x 4th derivative: cos x ; What??? That's the original equation!! After taking the 4th derivative of cos x, you end up the the same equation you started with. Therefore, the derivatives after that are going to be repeats of the first 4. In other words, the 5th derivative will be the same as the 1st derivative, the 6th derivative will be the same as the 2nd derivative, the 7th derivative will be the same as the 3rd derivative, the 8th derivative will be the same as the 4th, and so on and so forth. -sin x: 1 5 etc... -cos x: 2 6 sin x: 3 7 cos x: 4 8 We have found our pattern: after every 4 derivatives, the values/equations repeat. That also means that after ever multiple of 4, you will get back to the original equation. -sin x: 1 5 .... 49 -cos x: 2 6 .... 50!!!! sin x: 3 7 .... 51 cos x: 4 8 12 16 20 .... 44 48 52 The 50th derivative of cos x will be -cos x. Without doing the table, you could have divided 50 by 4. When you divide by 4, the remainder will be either 1, 2, 3, or no remainder which is means that you're back with the original equation. Dividing 5 by 4 will give you a remainder of 2 which means that the 50th derivative is the same as the 2nd derivative (-cos x). Either method works. If you're more visual and need to see the numbers changing, the table method is better for you! :)
Solve for x in the following equation: y = 2x^2 - x - 1.
Seeing that this equation is a quadratic equation (given by the degree of 2), we know that we can solve this equation in 4 different ways. 1.) Graphing 2.) Factoring 3.) Using the Quadratic Formula 4.) Completing the Square You may have seen these methods in a different order. For the purpose of this lesson, I ordered them according to simplest/easiest to more complex/hardest. _________________________________________________ If your teacher permits you access to a graphing calculator, the easiest method to solving this problem would be to simply graph the equation and locate the points at which the graph crosses or touches the x-axis (the x-intercepts). The x-intercepts are the solution to this equation. Because this equation is a quadratic, it will produce a graph that looks like a giant U. As you should hopefully see from the graph, this quadratic has 2 x-intercepts (thus, 2 solutions): x=1 and x=-1/2. Though some students don't, I recommend that you check both solutions with the original equation. You do that by plugging in (substituting) each solution into the equation, and then simplifying. If you get 0, you can officially say that x-intercept is a solution to the equation. x=1; 2(1)^2 - (1) - 1 = 0 2(1) - 1 - 1 = 0 2 - 1 - 1 = 0 1 - 1 = 0 0 = 0 CHECK!! x=-1/2; 2(-1/2)^2 - (-1/2) - 1 = 0 2(1/4) - (-1/2) - 1 = 0 (1/2) - (-1/2) - 1 = 0 ; A negative and a negative gives you a positive. (1/2) + (1/2) - 1 = 0 1 - 1 = 0 0 = 0 CHECK!! And you're done! ____________________________________________ If you aren't fortunate enough to be able to use a graphing calculator for this problem, the next easiest method to solve this equation is to factor it. (Some of the students I tutor find factoring very difficult which is why I recommend graphing first when possible.) After factoring the equation, set it equal to zero. You should have this: (2x+1)(x-1) = 0 Then, set each quantity equal to zero to find your solutions. 2x + 1 = 0 ; Subtracting 1 on both sides. 2x = -1 ; Isolate x by undoing the multiplication with division (Divide by 2 on each side). x= -1/2 x - 1 = 0 ; Add 1 on both sides. x = 1 And you're done!! As you can see, this method gives you the same solutions as graphing the equation. Each method with produce the same solution. It all depends on which method you prefer or which method you have the time for in a timed sessions (an in-class test per say). _________________________________________________ If you can't use a graphing calculator and really, really have trouble with factoring, I recommend you try using the quadratic formula next. I recommend this method next because as long as you know the quadratic formula, the pathway to solving this equation is literally plug-and-chug. Quadratic Formula: x = [-b ± √(b^2 - 4ac)]/2a Referring back to the original equation y = 2x^2 - x - 1: a = 2 b = -1 c = -1 Those are the only values you need. Now that you've determined which is which, you simply substitute each value into the equation in place of their respected variable. Once solve, you will get the same solutions: x = -1/2 and x = 1. _______________________________________________ In my opinion, completing the square should be the absolute last result. Of the 4 methods you can use to solve a quadratic equation, it is in my opinion the hardest method simply because of its complexity compared to the others. I am only not showing it here because it is very difficult to explain in words versus a visual explanation, and I believe an attempt to do so may result in confusing you even more.