# Tutor profile: Alyssa J.

## Questions

### Subject: Pre-Algebra

If you divide by 3, and the answer is 75, what was your original number?

Let's call our original number "x". We can write an equation for x like this: x/3 = 75. We need to get x by itself on one side, so we need to "un-do" the division by multiplying both sides by 3. 3(x/3) = 3(75) x = 225 So our original number is 225.

### Subject: Calculus

A cable that weighs 2lbs/ft is attached to a bucket of water that weighs 800lbs. Initially, the bucket is at the bottom of a 500ft deep well. Calculate the amount of work to lift the bucket to the top of the well.

Before we set up the problem, we need to determine the force of the cable and the bucket at any height, which we will call "x", in the well. Total force = weight of bucket + weight of cable The weight (a type of force) of the bucket will always be 800lbs no matter what height it is at, but the cable's weight will vary depending on its height. The cable starts at a length of 500ft, so at height "x", the cable's length is 500 - x. Since the cable weighs 2lbs/ft, its weight at time "x" is 2(500 - x) 1000 - 2x Now, we can plug in these values for total force. Total force = 800 + (1000 - 2x) Total force = 800 + 1000 - 2x Total force = 1800 - 2x Now, we can find the work. The work done to bring the bucket up to the top of the well is equal to the integral of the total force from 0 to 500. W = \int_0^500 \! 1800 - 2x \, \mathrm{d}x W = 1800x - x^2 from 0 to 500 W = [1800(500) - (500)^2] - [1800(0) - (0)^2] W = 900000 - 250000 W = 650000 ft-lbs So it takes 650,000 ft-lbs to raise the 800lb bucket on a 2lbs/ft cable 500 feet up.

### Subject: Algebra

Your math teacher gives you a piece of cardboard and tells you that its area is 24 square inches (24 in^2). Your teacher also tells you that the width of the cardboard is 2 inches less than the length. From this information, find the length and width (in inches) of the piece of cardboard.

We know that the area (A) of a rectangle is its width (W) multiplied by its length (L). This gives us the equation W x L = A. We also know that the width is 2 inches shorter than the length, so we can write the equation W = L - 2. Now, we can solve the first equation we found ( W x L = A ) by plugging in our second equation (W = L - 2). W x L = A (L - 2) x L = A (L x L) - (2 x L) = A L^2 - 2L = A The last piece of information we have is that the area of the piece of cardboard is 24 square inches, so we can plug that in for A. L^2 - 2L = A L^2 - 2L = 24 Now we move all terms to one side so that we can factor it. L^2 - 2L = 24 L^2 - 2L - 24 = 0 (L - 6) (L + 4) = 0 This tells us that L equals either 6 or -4. Since a rectangle can't have a negative length or width, we ignore the -4. Now that we've found L, we can plug it into the equations above for length and width. For length: L = 6 For width: W = L - 2 W = 6 - 2 W = 4 So the length of the piece of cardboard is 6 inches, and the width is 4 inches.

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