a. When the ball thrown vertically up ward to reach its maximum height at velocity = 0 then complete its motion vertically down ward to strike the ground its displacement = zero

∴ s(t) = 0 ⇒ $80t-16t^{2}=0$

∴ $16t(5 - t) = 0$

5 - t = 0⇒ ∴ t = 5 seconds

b. To tind the time that the ball will be over 96 ft over the ground substitute in $s(t)=80t-16t^{2}$

$96=80t-16t^{2}$

$16t^{2}-80t+96=0$

$16(t^{2}-5t+6)=0$

$16(t-3)(t-2)=0$

t = 3 sec. and t = 2 sec.

The ball will be at height 96 ft at time 2 sec. and 3 sec.

So the ball will be at height over than 96 = 3 - 2 = 1 sec.

8 0

2 years ago

State whether the value for the unknown makes inequalities <br><br> 74/e < 12

Nezavi [6.7K]

Answer:

Yes

Step-by-step explanation:

74/e < 12

Since e is in the denominator, e can not be 0.

If e > 0,

74/e < 12

74 < 12e

e > 74/12

e < 37/6

If e < 0,

(When you multiply by e, which is a negative value, inequality reverses)

74 > 12e

e < 74/12

e < 37/6

6 0

2 years ago

Read 2 more answers

Jeremy has a box of 500 nails that weighs 1.35 kilograms. He uses 60 nails to build a birdhouse. How much do the nails in the bi

madreJ [45]

.162 kilograms?

How I got my answer:

1.35÷500=.0027

.0027×60=.162

8 0

3 years ago