EXAMPLE 6 A projectile is fired with muzzle speed 170 m/s and an angle of elevation 45° from a position 20 m above ground level. Where does the projectile hit the ground and with what speed? SOLUTION If we place the origin at ground level, then the initial position of the projectile is (0, 20) and so we need to adjust the parametric equations of the trajectory by adding 20 to the expression for y. With v0 = 170 m/s, α = 45°, and g = 9.8 m/s2, we have

step-by-step explanation:

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.

ur right ive taken the test

step-by-step explanation: