NO LINKS OR FILES ATTACHED OR I WILL HAVE YOU REPORTED I JUST WANT HELP SO NO BS PLS You throw a baseball from 6 feet above the ground with an initial vertical velocity of 38 feet per second. Your friend catches the ball at a height of 2 feet. The equation below models the height of the ball as a function of time. HOW LONG IS THE BALL IN THE AIR? (Round to the nearest hundredth of a second,
y=-16x^(2)+38x+6
You throw a baseball from 6 feet above the ground with an initial vertical velocity of 38 feet per second. Your friend catches the ball at a height of 2 feet. The equation below models the height of the ball as a function of time. FIND THE MAXIMUM HEIGHT OF THE BALL? (Round to the nearest tenth of a foot
y=-16x^(2)+38x+6

65

step-by-step explanation:

14

step-by-step explanation:

14 x 3 = 42

p ÷ r = tan(x)

step-by-step explanation:

given that there is a figure below shows a right triangle: a right triangle is shown with hypotenuse equal to q units and the length of the legs equal to p units and r units.

graph is missing so i will create the graph as per given information.

the angle between the legs having lengths p units and q units is y degrees. the angle between the legs having lengths r units and q units is x degrees.

now we need to find about what is p ÷ r equal to.

so we can use trigonometric ratio tan(x)=[altitude]/[base] to find the value of p ÷ r, because p is the altitude and r is the base.

p ÷ r=tan(x)

hence final answer is p ÷ r=tan(x).

e^-kt = n/n0 = 0.79  

-kt = ln 0.79  

t = - ln 0.79/0.0001  

= 2357 years

step-by-step explanation: