My question is hat if an Air force combat training jet execute, a high-speed maneuver along the a path 2ysquare minus x square that that is equal to 8. How close does the jet come to a town located at (3,0)? Let m denote the square of the distance from a point (x, y) on the path to (3,0) and finding the minimum value of m. What is square root m?

Step-by-step explanation:

Let the function be:

f(x, y) = 2(y^2) - x^2 = 8

Since the shortest distance is basically a straight line from point (3, 0) to

f(3, y).

2y^2 - 3^2 = 8

2y^2 = 17

y = sqrt(17/2)

So the nearest point on the planes course to the point (3,0) is (3, sqrt(17/2))

Now lets use the distance formula:

x1 = 3

y1 = 0

x2 = 3

y2 = sqrt(17/2)

So m in this case is just 17/2 and the square root of m is sqrt(17/2) = 2.91547594742

this may not be at all, but i use the same website you're using for school, and if i'm stuck doing one of these i try to look for it on the section.

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.