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Stephen G.

Clemson Athletics Tutor

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Applied Mathematics

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Question:

Fast-Build construction company noticed it takes 3 people about 4 days to build 10 houses. How long would it take 6 people to build 5 houses?

Stephen G.

Answer:

To solve problems like this it is important to notice the ratio of changes that are occurring. Notice that the number of workers is doubling, so the amount of time required will decrease by a factor of two. Also notice, the number of houses being made is decreasing by a factor of two, so the time it takes will also decrease by a factor of two. Combining the two facts together, the time will decrease by a factor of 4. The total time for 6 people to build 5 houses will be 1 day

Calculus

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Question:

$$ \int \cos(x)\sin(x) \mathrm{d}x $$

Stephen G.

Answer:

This problem is a u-substitution problem; with practice, it becomes easier to detect that this problem is a u-substitution problem. $$ u = \sin(x) $$ $$ \mathrm{d} u = \cos(x) \mathrm{d}x $$ For this problem $$ u$$ can be chosen to be either $$ \sin(x) \mathrm{or} \cos(x)$$, however $$\sin(x)$$ is slightly simpler. Notice all three components of the intergral are present in $$ u $$ and $$\mathrm{d} u$$ ($$\cos(x), \sin(x), \mathrm{d}x$$). Now substitute in $$u$$ and $$\mathrm{d}u$$, giving: $$\int u\mathrm{d}u$$, which is a simpler integral to evaluate. $$\int u\mathrm{d}u = \frac{u^{2}}{2} + C$$; DO NOT FORGET $$ +C$$, it is the most common mistake for for basic calculus problems Remember you must change all "u's" back into "x's", so the final answer is: $$\frac{\sin(x)^{2}}{2} + C$$

Algebra

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Question:

The width of Clemson University's football field is 3/4 the length of the field and the perimeter of the field is 140. What is the area of the field?

Stephen G.

Answer:

The above problem asks for the area of the field, so it is necessary to identify what is needed: A = L*W. From the first sentence, it can be noticed that the problem is a system of equations for length and width: W = 3/4 * L; "The width of the field is 3/4 the length" 2*W + 2*L = 140; "The perimeter of the field is 140" (remember perimeter is the sum of the sides) This system can be solved by direct substitution. Substituting for W on the second equation: 2*(3/4)*L + 2*L = 140 3/2 *L +2*L = 140; simplifying the fractions on the first term 7/2 * L = 140; combining like terms( remember to add fractions denominator must be the same, so 2 becomes 4/2) L = 280/7 = 40; multiply by 2/7 on both sides to cancel the 7/2. Now that L is known, W can be solved for: W = 3/4 L W = 30; 3/4 * 40 ( 120/4 = 30) Now remember the problem asked for the area: A = L*W A = 30*40 = 1200

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