She put out two more recovery tests smh

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$

Since time can only be positive we reject the $t_{2}$ solution and we keep that Ben's book took $t=4.3276$ seconds to reach the ground.

Similarly solving for Josh we obtain

$t_{1}=3.6794sec\\t_{2}=-0.6794sec$

Thus again we reject the negative and keep the positive solution, so Josh's book took $t=3.6794$ seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

$t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482$

So to answer the original question:

Whose textbook reaches the ground first and by how many seconds?

Josh's textbook reached the ground first
Josh's textbook reached the ground first by a difference of $t=0.6482$

3 0

3 years ago

Answer correct for brainliest

Illusion [34]

The answer is 59/12

4 0

2 years ago

Read 2 more answers

The sum of two numbers is 28. one number is 14 more. find the numbers

ruslelena [56]

The numbers are: "7 " and "21 " .
_______________________________________
Explanation:
_______________________________________
The numbers are: "x" and "x + 14" .

x + (x + 14) = 28 . Solve for "x" ; and then solve for "x + 14" .
_______________________________________
→ x + (x + 14) = 28 ;

Rewrite as:

→ x + x + 14 = 28 ;

→ 1x + 1x + 14 = 28 ;

→ 2x + 14 = 28 ;

Subtract "14" from each side of the equation;

→ 2x + 14 − 14 = 28 − 14 ;

 → 2x = 14 ;

Divide EACH SIDE of the equation by "2" ;
to isolate "x" on one side of the equation; & to solve for "x" ;

→ 2x / 2 = 14 / 2 ;

→ x = 7 .
_________________________________________
So; one of the numbers is: " 7 " .

The other number is: " x + 14 " ; which equals: " 7 + 14 = 21".

The other number is: "21 " .
_________________________________________
The numbers are: "7 " and "21 " .
_________________________________________

5 0

3 years ago

Debra walked her dog to the south. They walked 50 meters in 150 seconds.

abruzzese [7]

Answer:

B

Step-by-step explanation:

Because... 150 divided by 50 = 3

:)

4 0

2 years ago

The graph above is a transformation of the function x2 Write an equation for the function graphed above

lutik1710 [3]

Answer: y=0,5x²-x-1,5.

Step-by-step explanation:

$(-1;0)\ \ \ \ (3;0)\ \ \ \ (1;-2) \ \ \ \ y=ax^2+bx+c\ ?\\\left\{\begin{array}{ccc}a*(-1)^2+b*(-1)+c=0\\a*3^2+b*3+c=0\\a*1^2+b*1+c=-2\end{array}\right \ \ \ \ \ \left\{\begin{array}{ccc}a-b+c=0\ \ (1)\\9a+3b+c=0\ (2)\\a+b+c=-2\ (3)\end{array}\right \\$