11. a carousel has a diameter of 50 feet. to the nearest foot, how far does a child

seated near the outer edge travel when the carousel makes 8

revolutions?

1256 ft

Step-by-step explanation:

The circumference of a circle tell us how long is the line which enclose the circle. So, after 1 revolution a child  seated near the outer edge travels a distance equal to the circumference of the carousel. If 8 revolutions are made the distance travelled will be 8 times the circumference.

Circumference = π · diameter = π · 50 ft = 157 ft

Distance travelled  = 8 · 157 ft = 1256 ft

since the airplane is moving with constant speed then the distance travelled can be calculated by x=u*t

in km/h would be:

x=640*3.5=2.240 km

in m/s would be:

u=640 km/h= 640.000 m/h =640.000/ 3600 m/s =177.8 m/s

t= 3.5 h= 3.5 * 3600 s=12.600 s

then x=u*t= 177.8*12.600= 2.240.280 m

answer

see attachment

explanation

the given function is

this function has y-intercept at

-4 and a slope of 2.

note that, the graph has a positive slope.

the graph that represents this function is the straight line that moves upward from left to right.

the graph touches the y-axis at -4.

the correct choice is the graph shown in the attachment.