So is doubling, first off, meaning, if the current amount say P, then when it doubles is 2P, or double that, or P + P, so the rate of growth is 100%, since it's doubling.
and is doing it every 9 hour cycle, thus
![\bf \textit{Periodic/Cyclical Exponential Growth}\\\\ A=P(1 + r)^{\frac{t}{c}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\to &650\\ r=rate\to 100\%\to \frac{100}{100}\to &1.00\\ t=\textit{elapsed time}\to &10\\ c=period\to &9 \end{cases} \\\\\\ A=650(1 + 1)^{\frac{10}{9}}\implies A=650(2)^{\frac{10}{9}}\implies A=650\sqrt[9]{2^{10}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7BPeriodic%2FCyclical%20Exponential%20Growth%7D%5C%5C%5C%5C%0AA%3DP%281%20%2B%20r%29%5E%7B%5Cfrac%7Bt%7D%7Bc%7D%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0AA%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%0AP%3D%5Ctextit%7Binitial%20amount%7D%5Cto%20%26650%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%2610%5C%5C%0Ac%3Dperiod%5Cto%20%269%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D650%281%20%2B%201%29%5E%7B%5Cfrac%7B10%7D%7B9%7D%7D%5Cimplies%20A%3D650%282%29%5E%7B%5Cfrac%7B10%7D%7B9%7D%7D%5Cimplies%20A%3D650%5Csqrt%5B9%5D%7B2%5E%7B10%7D%7D)
The coordinates of the midpoint if the endpoints are (x₁,y₁) and (x₂,y₂):
One endpoint is (3,5), the midpoint is (5,-11). The other endpoint is (x,y).
The coordinates of endpoint M are (7,-27).
The y axis was the axis the point was reflected over
I think its 52.3 yards but im not sure
Answer:
c)
Step-by-step explanation:
To solve this, remember some properties of trigonometric functions.
Let 
. Then, by definition of inverse function, 
. Multiply by -1 to both sides of this equation to get 
.
Note that 
because sine is an odd function and cosine is an even function. Then 
. Take the inverse tangent in both sides to get 
.
Using the previous equations, we obtain:
