Given the two functions:
![\begin{gathered} R(x)=2\sqrt[]{x} \\ S(x)=\sqrt[]{x} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20R%28x%29%3D2%5Csqrt%5B%5D%7Bx%7D%20%5C%5C%20S%28x%29%3D%5Csqrt%5B%5D%7Bx%7D%20%5Cend%7Bgathered%7D)
We need to find (RoS)(4). THis is the functional composition. We take S(x) and put it into R(x) and then put "4" into that composed function. Shown below is the process:
![(RoS)(x)=2\sqrt[]{\sqrt[]{x}}](https://tex.z-dn.net/?f=%28RoS%29%28x%29%3D2%5Csqrt%5B%5D%7B%5Csqrt%5B%5D%7Bx%7D%7D)
When we plug in "4", into "x", we have:
There's a simple formula for the area of an ellipse: pi*a*b, where a and b are , half the lengths of the long axis and the short axis respectively.
Here, A = pi*(12 ft)(4 ft) = 48 pi ft^2, or about 150.8 ft^2.
Answer:
D - ASA
This should be the answer since FH and IG can be said to be parallel and therefore we can find both alternate angles or a vertical opposite angle with one side given. We know that ΔFHJ is reduced by a factor of 2 to get ΔGIJ.
if those r coordinates you can use the Pythagorean the to find the distance
plot the point and create a right triangle with the origin
could up the spaces and depending on the number, subtract the side from the hypotenuse or add the sides to find the hypotenuse
then square root both sides
Answer:
Step-by-step explanation: