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If the population is doubling every 10 years, the rate of change is 100%, for the period of 10 years, so, whatever it happens to be at the time, it grows by 100%, namely it doubles.
![\bf \textit{Periodic Exponential Growth}\\\\ A=I(1 + r)^{\frac{t}{p}}\qquad \begin{cases} A=\textit{accumulated amount}\\ I=\textit{initial amount}\to &4100\\ r=rate\to 100\%\to \frac{100}{100}\to &1.00\\ t=\textit{elapsed time}\to &8\\ p=period\to &10 \end{cases} \\\\\\ A=4100(1 + 1)^{\frac{8}{10}}\implies A=4100(2)^{\frac{4}{5}}\implies A=4100\sqrt[5]{2^4}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7BPeriodic%20Exponential%20Growth%7D%5C%5C%5C%5C%0AA%3DI%281%20%2B%20r%29%5E%7B%5Cfrac%7Bt%7D%7Bp%7D%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0AA%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%0AI%3D%5Ctextit%7Binitial%20amount%7D%5Cto%20%264100%5C%5C%0Ar%3Drate%5Cto%20100%5C%25%5Cto%20%5Cfrac%7B100%7D%7B100%7D%5Cto%20%261.00%5C%5C%0At%3D%5Ctextit%7Belapsed%20time%7D%5Cto%20%268%5C%5C%0Ap%3Dperiod%5Cto%20%2610%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D4100%281%20%2B%201%29%5E%7B%5Cfrac%7B8%7D%7B10%7D%7D%5Cimplies%20A%3D4100%282%29%5E%7B%5Cfrac%7B4%7D%7B5%7D%7D%5Cimplies%20A%3D4100%5Csqrt%5B5%5D%7B2%5E4%7D)
Answer:
7) K'(1, 1), N'(4, 0), Q'(4, 4)
9) F'(4, 2), W'(4, 5), K'(3, 2)
Step-by-step explanation:
When you are not sure about reflections and rotations, it can be useful to actually draw a graph. The line of reflection is the perpendicular bisector of the segment connecting a point with its image. That is, the image is the same distance from the line on the other side. A point that is <em>on</em> the line of reflection is "invariant," it doesn't move.
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7) Reflecting across the x-axis changes the sign of the y-coordinate. That is all. Any point on the x-axis (y=0) is unchanged. The transformation is ...
(x, y) ⇒ (x, -y)
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9) Reflecting across a vertical line changes the x-coordinate. The new value of x will be ...
x' = 6 - x
The "6" is double the x-coordinate of the vertical line. The transformation is ...
(x, y) ⇒ (6-x, y)
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<em>Derivation of the formula for reflection on x=3</em>
In general if M is the midpoint of a point (A) and its reflection (A'), then we have ...
M = (A +A')/2
2M = A + A'
2M -A = A'
When a point is reflected across a vertical line, all of the y-coordinate values remain unchanged. So, for A=(x, y) and M=(3, y), we have ...
A' = 2(3, y) -(x, y) = (6-x, 2y-y)
A' = (6 -x, y)
So, the reflection transformation across the line x = 3 is ...
(x, y) ⇒ (6-x, y)
