An astronaut on the moon throws a baseball upward. the astronaut is 6​ ft, 6 in.​ tall, and the initial velocity of the ball is 50 ft per sec. the height s of the ball in feet is given by the equation s=-2.7t^2+50t+6.5 ​, where t is the number of seconds after the ball was thrown. complete parts a and b. A) After how many seconds is the ball 10 ft above the​ moon's surface?

B)How many seconds will it take for the ball to hit the​ moon's surface?

Step-by-step explanation:

The position function is

and if we are looking for the time(s) that the ball is 10 feet above the surface of the moon, we sub in a 10 for s(t) and solve for t:

and

and factor that however you are currently factoring quadratics in class to get

t = .07 sec and t = 18.45 sec

There are 2 times that the ball passes 10 feet above the surface of the moon, once going up (.07 sec) and then again coming down (18.45 sec).

For part B, we are looking for the time that the ball lands on the surface of the moon. Set the height equal to 0 because the height of something ON the ground is 0:

and factor that to get

t = -.129 sec and t = 18.65 sec

Since time can NEVER be negative, we know that it takes 18.65 seconds after launch for the ball to land on the surface of the moon.

c.

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.

there are 70 more legs than eyes at the dog show (196 – 126 = 70). dogs have two more legs than eyes, and humans have the same number of legs as eyes. thus 70 is twice the number of dogs at the show, so the number of dogs is half of 70, which is 35. to get the number of humans, subtract 35 from 63: 63 – 35 = 28.

step-by-step explanation: