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2 years ago

13

!!!!________________

2 answers:

myrzilka [38]2 years ago

8 0

!!!!!!!!!!!!!!!! It’s b

pychu [463]2 years ago

5 0

Answer:

B

Step-by-step explanation:

The answer is B as it is the only one in which the shape at the bottom remains the cross section shape for the object. For example in a sphere the cross section near the bottom wouldn't be the same as the one at the middle. The only shape that does have the same cross section all the time is the cuboid so it is the answer

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2 years ago

2. When Joey dives off a diving board, the equation of his pathway can be modeled by h = -16- + 15 + 12.

Sergeu [11.5K]

Answer:

a) The maximum height is approx 15.5 unit.

b) The time it will take for Joey to reach the water is 1.45 hour.

Step-by-step explanation:

Given : When Joey dives off a diving board, the equation of his pathway can be modeled by $h(t)= -16t^2+15t + 12$

To find : a) Find Joey's maximum height.

b) Find the time it will take for Joey to reach the water.

Solution :

Modeled $h(t)= -16t^2+15t + 12$ ....(1)

a) To find maximum height

Derivate (1) w.r.t. t,

$h'= -32t+15$

For critical point put h'=0,

$-32t+15=0$

$t=\frac{15}{32}$

$t=0.46875$

Derivate again w.r.t. t,

$h''= -32$

It is maximum at t=0.46875.

Substitute t in equation (1),

$h(0.46875)= -16(0.46875)^2+15(0.46875)+12$

$h(0.46875)= -3.515625+7.03125+12$

$h(0.46875)= 15.515625$

The maximum height is approx 15.5 unit.

b) To find the time it will take for Joey to reach the water.

Put h=0 in equation (1),

$-16t^2+15t + 12=0$

Apply quadratic formula, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here, a=-16 , b=15, c=12

$t=\frac{-15\pm\sqrt{(15)^2-4(-16)(12)}}{2(-16)}$

$t=\frac{-15\pm\sqrt{225+768}}{-32}$

$t=\frac{-15\pm\sqrt{993}}{-32}$

$t=\frac{-15+\sqrt{993}}{-32},\frac{-15-\sqrt{993}}{-32}$

$t=−0.515,1.453$

Reject negative value.

The time is t=1.45.

The time it will take for Joey to reach the water is 1.45 hour.

5 0

3 years ago