A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find out the time at which the rocket will reach its max, to the nearest 100th of a second. y=-16x^2+161x+108
y=−16x
2
+161x+108

5.03secs

Step-by-step explanation:

Given the height reached by the rocket expressed as;

y = -16x^2 + 161x + 108

The rocket reaches it max height when its velocity is zero

Velocity is expressed as;

V(x) = dy/dx  = 0

dy/dx = -32x + 161

0 = -32x + 161

32x = 161

x = 161/32

x = 5.03

Hence it will take 5.03secs for the rocket to reach its maximum height

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

1/3 + 1/5 = 8/15.

step-by-step explanation:

hope it