Hurry answer this and i wil give you
malia solved an equation, as shown below:

step 1: 6x = 48
step 2: x = 48 – 6
step 3: x = 42

part a: is malia's solution correct or incorrect? if the solution is incorrect, explain why it is incorrect and show the correct steps to solve the equation.

part b: how many solutions will this equation have?

her resolution is incorrect because it's 6 times x=48. so to get rid of of the six you have to divide it on both sides, so x can be alone. and that's will only have one solution which is x=8

a) The distance from the television camera to the rocket is changing at that moment at a speed of

600 ft/s

b) the camera's angle of elevation is changing at that same moment at a rate of

0.16 rad/s = 9.16°/s

Step-by-step explanation:

This is a trigonometry relation type of problem.

An image of when the rocket is 3000 ft from the ground is presented in the attached image.

Let the angle of elevation be θ

The height of the rocket at any time = h

The distance from the camera to the rocket = d

a) At any time, d, h and the initial distance from the camera to the rocket can be related using the Pythagoras theorem.

d² = h² + 4000²

Take the time derivative of both sides

(d/dt) (d²) = (d/dt) [h² + 4000²]

2d (dd/dt) = 2h (dh/dt) + 0

At a particular instant,

h = 3000 ft,

(dh/dt) = 1000 ft/s

d can be obtained using the same Pythagoras theorem

d² = h² + 4000² (but h = 3000 ft)

d² = 3000² + 4000²

d = 5000 ft

2d (dd/dt) = 2h (dh/dt) + 0

(dd/dt) = (h/d) × (dh/dt)

(dd/dt) = (3000/5000) × (1000)

(dd/dt) = 600 ft/s

b) If the television camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing at that same moment?

At any moment in time, θ, h and the initial distance of the camera from the base of the rocket are related through the trigonometric relation

Tan θ = (h/4000) = 0.00025h

Taking the time derivative of both sides

(d/dt) (Tan θ) = (d/dt) (0.00025h)

(Sec² θ) (dθ/dt) = 0.00025 (dh/dt)

At the point where h = 3000 ft, we can calculate the corresponding θ at that point

Tan θ = (3000/4000)

θ = tan⁻¹ (0.75) = 0.6435 rad

(Sec² θ) (dθ/dt) = 0.00025 (dh/dt)

(Sec² 0.6435) (dθ/dt) = 0.00025 (1000)

1.5625 (dθ/dt) = 0.25

(dθ/dt) = (0.25/1.5625) = 0.16 rad/s

Hope this Helps!!!