1.
After each washing, \(1\%\) of the dye in blue jeans is washed out. How much of the original dye remains after \(10\) washings?
Step 1. If \(1\%\) of the dye is washed out each time, then \(99\%\) remains after each washing. \[ r=0.99 \] Step 2. Let the original amount of dye be \(1\). \[ a=1 \] Step 3. After \(10\) washings, the remaining amount is: \[ 1(0.99)^{10} \] Step 4. Calculate. \[ (0.99)^{10}\approx 0.904 \] Step 5. Convert to a percentage. \[ 0.904\approx 90\% \] Therefore, about \(90\%\) of the original dye remains. Answer: A
2.
Determine the common ratio of the following geometric sequence: \(1,-\frac{1}{3},\frac{1}{9},-\frac{1}{27},\cdots\)
Step 1. The common ratio of a geometric sequence is found by dividing a term by the previous term. \[ r=\frac{t_2}{t_1} \] Step 2. Use the first two terms \(t_1=1\) and \(t_2=-\frac{1}{3}\). \[ r=\frac{-\frac{1}{3}}{1}=-\frac{1}{3} \] Step 3. Check with the next terms. \[ \frac{\frac{1}{9}}{-\frac{1}{3}}=-\frac{1}{3},\quad \frac{-\frac{1}{27}}{\frac{1}{9}}=-\frac{1}{3} \] Therefore, the common ratio is \(-\frac{1}{3}\). Answer: D
3.
Find the first term for the following series: \(\sum_{k=1}^{7}3(2)^{k-1}\)
Step 1. The first term occurs when the index starts at \(k=1\). Step 2. Substitute \(k=1\) into the expression \(3(2)^{k-1}\). \[ 3(2)^{k-1}=3(2)^{1-1} \] Step 3. Simplify. \[ 3(2)^0=3(1)=3 \] Therefore, the first term is \(3\). Answer: E
4.
A ball is dropped from a height of \(16\text{ m}\). After each bounce the ball rises to \(75\%\) of its previous height. Determine the total vertical distance the ball has traveled when it bounced for the \(10\)th time.
Step 1. The ball first falls \(16\text{ m}\). Step 2. After each bounce, the height is \(75\%\) of the previous height, so the common ratio is: \[ r=0.75 \] Step 3. The first bounce height is: \[ a=16(0.75)=12 \] Step 4. When the ball has bounced for the \(10\)th time, there are \(9\) full up-and-down bounce heights after the initial drop, as shown in the solution. Step 5. Use the finite geometric sum formula for the bounce heights. \[ S_9=\frac{a(1-r^9)}{1-r} \] Step 6. Substitute \(a=12\) and \(r=0.75\). \[ S_9=\frac{12(1-0.75^9)}{1-0.75} \] Step 7. The total vertical distance is the initial drop plus twice the bounce-height sum. \[ \begin{aligned} \text{Total distance}&=16+2S_9\\&=16+2\left(\frac{12(1-0.75^9)}{1-0.75}\right)\\&\approx 104.79\end{aligned} \] Therefore, the total vertical distance is approximately \(104.79\text{ m}\). Answer: A
5.
Find the exact value of the sum for the following series: \(\sum_{i=2}^{\infty}3\left(\frac{2}{3}\right)^i\)
Step 1. This is an infinite geometric series. \[ \sum_{i=2}^{\infty}3\left(\frac{2}{3}\right)^i \] Step 2. The common ratio is: \[ r=\frac{2}{3} \] Step 3. Find the first term of the series by substituting \(i=2\). \[ a=3\left(\frac{2}{3}\right)^2=3\left(\frac{4}{9}\right)=\frac{4}{3} \] Step 4. Use the infinite geometric series formula. \[ S=\frac{a}{1-r} \] Step 5. Substitute \(a=\frac{4}{3}\) and \(r=\frac{2}{3}\). \[ \begin{aligned} S&=\frac{\frac{4}{3}}{1-\frac{2}{3}}\\&=\frac{\frac{4}{3}}{\frac{1}{3}}\\&=\frac{4}{3}\times 3\\&=4\end{aligned} \] Therefore, the exact sum is \(4\). Answer: B
6.
If the sum of an infinite geometric series is \(64\) and the first term is \(8\), determine the common ratio.
Step 1. Use the infinite geometric series formula. \[ S_{\infty}=\frac{a}{1-r} \] Step 2. Substitute \(S_{\infty}=64\) and \(a=8\). \[ 64=\frac{8}{1-r} \] Step 3. Solve for \(r\). \[ \begin{aligned} 64(1-r)&=8\\64-64r&=8\\-64r&=8-64\\-64r&=-56\\r&=\frac{-56}{-64}\\r&=\frac{7}{8}\end{aligned} \] Therefore, the common ratio is \(\frac{7}{8}\). Answer: B
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