Aball is launched into the sky at 272 feet per second from the roof of a skyscraper 1,344 feet tall. the equation for the ball’s height h at time t seconds is h = -16t2 + 272t + 1344. when will the ball strike the ground?

Time taken to reach the ball to the ground is 21 seconds.                  

Step-by-step explanation:

Given : A ball is launched into the sky at 272 feet per second from the roof of a skyscraper 1,344 feet tall. The equation for the ball’s height h at time t seconds is .

To find : When will the ball strike the ground?

Solution :

The equation for the ball's height 'h' at time 't' is given by,

When the ball strike the ground the height of the ball became zero.

Substitute h=0 in the given equation,

Taking 16 common,

or

Solve by middle term split,

Reject t=-4 as time can never be negative.

Therefore, Time taken to reach the ball to the ground is 21 seconds.

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

do it yourself this is easy

step-by-step explanation: