An airplane flys at a constant altitude of 2 miles and a constant speed of 600 miles per hour on a straight course that will take it directly over a kangaroo on the ground. how fast is the angle of elevation of the kangaroo's line of sight increasing when the distance from the kangaroo to the plane is 3 miles? give your answer in radians per minute.

Let us say that y = vertical height and z = hypotenuse 
dz/dt = 600 
dy/dt = 0 since y (altitude) is constant 
sin(θ) = 2/3 
Calculating for the angle: θ = arcsin(2/3) = 0.729728 radians 
sin(θ) = y/z 
z sin(θ) = y 
We then differentiate the two sides with respect to time t:
dz/dt sin(θ)+ z cos(θ) dθ/dt = dy/dt 
(600) (2/3) + (3) cos (0.729728) dθ/dt = 0 
400 + [3 / cos (0.729728) ] dθ/dt = 0 
400+ 4.0249 dθ/dt = 0 
dθ/dt = -98.38 radians/hour 
= -99.38/60 radians / min 
= -1.66 radians / min = 2PI - 1.66 radians / min 
= 4.62 radians / min    (ANSWER)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.

explanation:

step-by-step explanation: