The flying squirrel of eastern Texas can drop a height of 60 feet. A function that approximates this drop is h=-16t^2+60, where h is the height in feet and t is the time in seconds. About how many seconds does it take for the squirrel to drop 60 feet?

About 1.94 seconds

Step-by-step explanation:

The quadratic equation gives the height of the squirrel from its starting position, which is 60 feet above ground.

The equation is:

To find time it takes to "drop" 60 feet, means the ground level, where height is equal to 0.

Or, h = 0

We substitute h = 0 into the equation and solve for t:

So, it takes the squirrel about 1.94 seconds to drop 60 feet

part a

x     y

0     8

1     7.5

2     7

3     6.5

4     6

5     5.5  

for every hour i deducted .5 inches from the length of the candle

part b

it is a function because for every x value there is only one y value

part   c

for every x value there is only one y value

x       y

0       8

1       7.6

2       7.2

3       6.8

step-by-step explanation:

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."