An astronaut drops a rock into a crater on the moon. the distance, d(t), in meters, the rock travels after t seconds can be modeled by the function d(t)=0.8t^2 . what is the average speed, in meters per second, of the rock between 5 and 10 seconds after it was dropped?

Plug in 5 and 10 to the equation.

For the first one would be d(5)=0.8(5)^2. Multiply the equation, d(5)=4^2. Then you just multiply the exponent and the answer would be, d(5)= 10

The second one, you just do the same thing and the answer would be d(10)= 64

12m/s

Step-by-step explanation:

The distance is given by the function:

the distance at 5 seconds is:

and the distance at 10 seconds:

So, the distance covered in the interval of time between 5 and 10 seconds is:

All of this is done to find the average speed with the formula:

and we substitute the distance covered wich is 60m, and the time between 5 and 10 seconds which is 5 seconds.

the average speed is:

x > 1.3

step by step:

8.5−0.9x> 9.8−1.9x

step 1: simplify both sides of the inequality.

−0.9x+8.5> −1.9x+9.8

step 2: add 1.9x to both sides.

−0.9x+8.5+1.9x> −1.9x+9.8+1.9x

1x+8.5> 9.8

step 3: subtract 8.5 from both sides.

1x+8.5−8.5> 9.8−8.5

1x> 1.3

step 4: divide both sides by 1.

1x/1 > 1.3/1

the first one

step-by-step explanation:

just id the test.