Quiz 2 (Solution)

Date: MATH 1090-4 - Fall 2003

Note: Unless stated otherwise, answers without justification receive no credit. Please, show your work!

1. Compute and simplify the following expressions so that only positive exponents remain.
   1. $\frac{3x^{-2}}{(3x)^2} = \frac{3x^{-2}}{3^2 x^2} = 3^{1-2} x^{-2-2} = 3^{-1} x^{-4} = \frac{1}{3x^4}$.
   2. $\frac{3x^{-2}}{3x^2}=3^{1-1} x^{-2-2} = x^{-4} = \frac{1}{x^4}$.
   3. $\frac{(3x)^{-2}}{(3x)^2} = \frac{3^{-2} x^{-2}}{3^2 x^2} = 3^{-2-2} x^{-2-2} = 3^{-4}x^{-4}= \frac{1}{3^4x^4}$.

2. Solve the following equation. Be sure to check the solution in the original equation.

   $$x+\frac{1}{3} =2(x-\frac{2}{3})-6x$$

   
   

   We solve for $x$ as usual:

   $$\begin{split}x+\frac{1}{3} &=2(x-\frac{2}{3})-6x\\ x+\frac{1}... ...frac{1}{3}\\ 5x &= -\frac{5}{3}\\ x &= -\frac{1}{3} \end{split}$$

   
   

   Next, we check by plugging in $x=-\frac{1}{3}$. On the left hand side we have: $x+\frac{1}{3} = -\frac{1}{3} + \frac{1}{3} = 0$, while on the right hand side: $2(x-\frac{2}{3})-6x = 2(-\frac{1}{3} -\frac{2}{3}) - 6(-\frac{1}{3}) = 2(-\frac{3}{3}) + \frac{6}{3} = -2 + 2 =0$. Therefore, $x=-\frac{1}{3}$ is the solution of the given equation.

3. The dollars of interest, $I$, paid on a debt of $10000$ over $3$ years, when the interest rate is $r\%$ satisfies the equation

   $$\frac{I}{175.393} + 0.663 = r.$$

   
   
   If the interest rate is $19.8\%$, find the amount of interest paid during the $3$ years.

   We start by replacing $r=19.8$, so that $I$ satisfies

   $$\frac{I}{175.393} + 0.663 = 19.8.$$

   
   
   Then, we want to find the interest paid; that is, we solve for $I$:

   $$\begin{split}\frac{I}{175.393} + 0.663 &= 19.8\\ \frac{I}{175... ... 0.663\\ I &= 175.393 (19.8-0.663)\\ I &= 3356.496. \end{split}$$

   
   
   Thus, the interest paid over a period of three years is (rounded to two decimal digits) $ $3356.50$.