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# Tutor profile: Zach F.

Inactive
Zach F.
Math tutor for 5+ years
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## Questions

### Subject:Applied Mathematics

TutorMe
Question:

You want to buy a house for \$350,000.00. Being a savvy home-buyer, you want to put down at least 20% of the value as a down payment, to avoid having to to pay the extra cost of buying private mortgage insurance (PMI). If you put down the 20% down payment on a 30 year mortgage at 4.5%, what will your monthly mortgage payment amount be?

Inactive
Zach F.

To solve this, we can simply use the mortgage payment formula, which is as follows: $$principal*\frac{rate}{1-(1+rate)^{-total number of payments}}$$ In this case, we have principal = 270,000 (350,000 less our 20% down payment), rate = $$frac{.045}{12} = .00375$$, and our total number of payments is 30 years * 12 months/year = 360 payments. Plugging in, we get $$270000*\frac{.00375}{1-(1+.00375)^{-360}} = 1,368.05 /month$$

### Subject:Calculus

TutorMe
Question:

What is the slope of the tangent line to $$f\left( x \right) = 3{\left( {x^2 -5} \right)^2}$$ at $$x = 4$$?

Inactive
Zach F.

Using the chain rule, we have: $$f\left( x \right) = 2*3{\left( {x^2 -5} \right)}*2x = 12x^3-60x$$, since the derivative of $$x^2-5$$ is $$2x$$, and we must bring down the power from the quantity, reduce the power by one, and then multiply by the derivative of the inside. We then plug in $$x = 4$$ to obtain our solution--the slope of the tangent line at $$x = 4$$ is $$528$$.

### Subject:Algebra

TutorMe
Question:

What is a good way to think about the following problem? Solve the equation |x - 4| = 6.

Inactive
Zach F.

In my opinion, the best way to approach this problem is to start with absolute value. What is absolute value? If we draw a number line and compute the absolute value of several points, we will start to see that absolute value could be described as the distance from zero. This will also allow us to see that opposite-signed numbers have the same absolute value, and thus an absolute value equation will have two solutions. This is similar to 2nd-degree polynomials; we spend a lot of time using the quadratic formula to solve for two solutions for x. We can use an identical approach for this problem, if we square both sides first.

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