A ball is thrown into the air and its position is given by h ( t ) = − 6.3 t 2 + 53 t + 24 where h is the height of the ball in meters t seconds after it has been thrown. Find the maximum height reached by the ball and the time at which that happens.

Now we can set up the derivate equal to 0 and we have:

And solving for t we got:

For the second derivate respect the time we got:

So then we can conclude that t = 4.206 is a maximum for the function.

And the corresponding height would be:

So the maximum occurs at t = 4.206 s and with a height of 135.468 m

Step-by-step explanation:

For this case we have the following function:

In order to maximize this function we need to take the first derivate respect the time and we have:

Now we can set up the derivate equal to 0 and we have:

And solving for t we got:

For the second derivate respect the time we got:

So then we can conclude that t = 4.206 is a maximum for the function.

And the corresponding height would be:

So the maximum occurs at t = 4.206 s and with a height of 135.468 m

hi there!

question 1 answer: p/q can be terminating when q = 5^m * 2^n

question 2 answer: -1 and -1

question 3 answer: -4

question 4 answer: 5/12

hope this : )

system of linear equation

step-by-step explanation:

as given in the data

a= 20 , b= -12

let us suppose that it is a straight line equation

now we know the formula for the system of linear equation

b= ma + c )

here m is slope of a straight line which is equal to formula

m=   (ii)

and here c is a constant

now putting the value of a and b in equation (ii)

m=

m=

putting the value of a,b and m in equation (i) gives us

-12 = + c

-12=-12 + c

adding 12 to both sides

-12 +12 = -12 +12 +c

0 =c

or c=0

as the value of the constant is zero which gives us that it is a linear equation and this order pair satisfies the system of linear equations.