# Tutor profile: Ashlyn M.

## Questions

### Subject: Calculus

$$ \int_1^4 \mathrm{cos}{x} - \frac{3}{x^{5}} \mathrm{d}x $$

The first thing I would do is split the integral into two. This just helps me keep track of things. $$ \int_1^4 \mathrm{cos}{x} - \int_1^4\frac{3}{x^{5}} \mathrm{d}x $$. Then I would evaluate the first integral. The integral of cos is always sine. The first integral evaluates to cos(4) - cos(1). Now the second integral is a little trickier. I like to write fractions like that as multiplication $$3x^{-5}$$. To perform an integral on an expression like that you first subtract one from the exponent so it is $$3x^{-4}$$, then divide the expression by the number in the exponent: $$\frac{-3}{4x^{4}}$$. I combined two steps there, which was to also put the exponent back in the denominator and take away its negative sign. Now evaluate cos(4) - cos(1) - ($$\frac{-3}{4*4^{4}} - \frac{-3}{4*1^{4}}$$) with a calculator.

### Subject: Physics

A ball is dropped from the top of a 50m building. Determine the time it takes for the ball to hit the ground.

This is a kinematics problem so we are going to need the main kinematic equations on hand. The first thing to do in these types of problems is to write down the known variables then we can decide which equation to use. In this case the only one explicitly stated in the problem is the height, or d as represented in the kinematic equations. We can make some assumptions here as well. Since the ball is starting from rest, we know the initial velocity is 0. Since the ball is falling vertically, we know that the acceleration is just gravity, or 9.8m/s^2. So we have d, a, and v0 and we need t. The kinematic equation that contains all of these variables is d = v0t +1/2*(a*t^2). Now we plug in our known variables to get 50m = 0*t + 1/2 * (9.8m/s^2) * t^2 which reduces to 50m = 4.9m/s^2 * t^2. Divide both sides and get 10.2 s^2 = t ^2. take the square root and get t = 3.2s.

### Subject: Algebra

Teddy was selling tickets for the school play. He sold 10 more adult tickets than children tickets and he sold twice as many senior tickets as children tickets. Adult tickets cost $5, children's tickets cost $2, and senior tickets cost $3. Teddy made $700. How many children's tickets were sold for the play? How many adult tickets were sold? How many senior tickets were sold?

First thing is to write down how many unknowns are in the problem. In this case there are 3. Children's tickets, senior tickets, and adult tickets. There must be the same number of algebraic expressions as there are unknowns. The next step is to find these expressions as they are typically given to us in the problem. The question states 10 more adult tickets were sold than children's. Represented algebraically this is a = c +10. The problem also states that twice as many senior tickets were sold than children's which is s = 2c, algebraically. Finally the problem states "Adult tickets cost $5, children's tickets cost $2, and senior tickets cost $3. Teddy made $700", which we can represent as 5a + 2c + 3s = 700. Now we can use our first two expressions to solve the third by plugging in every variable as c. From the first expression we know c = a +10 and the second gives c = 2s or s = 2c. Now we can plug c in for s and a in the third expression to find 5(c+10) +2c + 3(2c) = 700. Multiplying out gives 5c + 50 + 2c + 6c = 700. Subtract the 50 to have the variables on one side and get 650 = 5c +2c+6c. Add and get 13c=650. divide both sides by 13 to get c = 50. Now plug c into our first two expressions and get a = 60 and s = 100. We can veryify by plugging these numbers into our third expression and get (3 * 100) + 2*50 + 5*60 = 300 +100 + 300 = 700!

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