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Sedbober [7]
3 years ago
15

What is the volume of a container that is 12 feet long, 4 feet wide and 8 feet high

Mathematics
2 answers:
Feliz [49]3 years ago
8 0

Answer:

384

Step-by-step explanation:

Andru [333]3 years ago
7 0

Answer:

384

Step-by-step explanation:

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What's the area of this rectangular prism?
Hitman42 [59]

Answer: 146cm²

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
An object launched straight up at a speed of 29.4 meters per second has a height, h, in meters of h , t seconds after the object
Nostrana [21]

Answer:

h = 44.06 meters (maximum height)

the time the object takes to complete this whole path is 6 seconds, this is why the time at which the object reaches its maximum height will between 0 and 6 seconds

Step-by-step explanation:

To solve this question, we need to first recognize that this is a constant acceleration problem, specifically, it can be thought of as a projectile motion problem.

Recall, the equations of motion:

1) v^2 - v_0^2 = 2a(s - s_0)\\2) s = v_0^2  + \frac{1}{2} at^2\\3) v = v_0 + at

What do we already know?

  • v_0 = 29.4 ms^-1
  • The launch is straight up
  • a = -9.81 ms^-2 this is the gravitational acceleration g
  • s_0 = 0 m, since our reference point is at s = 0, (the ground)

We can use use the Eq(1):

we know that when any object is launched up, at maximum height its velocity is going to be zero, v = 0 ms^-2

v^2 - v_0^2 = 2a(s - s_0)\\0^2 - (29.4)^2 = 2(-9.81)(s- 0)\\s = 44.06 m

this is the maximum height!

Why does t have to between zero and six?

We can answer this using a bit visualization, if you think about the second equation

s = v_0 t - \frac{1}{2}at^2\\ s = 29.4t - 4.905t^2

this is the equation of the whole trajectory that object makes.

and if you solve this by making s = 0, you will get the times at which the object was at the ground. the times will be 0s and 5.99s.

so the amount of time the object takes to go through this whole path is 6 seconds and this why the object will only reach its maximum height in between this time interval.

hope this helps :)

5 0
3 years ago
DOOOO NOTTT SKIPPPPPPP
Shalnov [3]
1st one RS=DE,
second one T=F,
last one R=D
7 0
3 years ago
Read 2 more answers
Darren wanted to see how contagious yawning can be. To better understand this, he conducted a social experiment in which he yawn
denis23 [38]

Answer:

10.5 minutes

Step-by-step explanation:

Thinking about the problem

The modeling function is of the form P(t)=A⋅Bf(t), where B=4B=4B, equals, 4 and f(t)=\dfrac{t}{10.5}f(t)=

10.5

t

​

f, left parenthesis, t, right parenthesis, equals, start fraction, t, divided by, 10, point, 5, end fraction.

Note that each time f(t)f(t)f, left parenthesis, t, right parenthesis increases by 111, the quantity is multiplied by B=4B=4B, equals, 4.

Therefore, we need to find the ttt-interval over which f(t)f(t)f, left parenthesis, t, right parenthesis increases by 111.

Hint #22 / 3

Finding the appropriate unit interval

fff is a linear function whose slope is \dfrac{1}{10.5}

10.5

1

​

start fraction, 1, divided by, 10, point, 5, end fraction.

This means that whenever ttt increases by \Delta tΔtdelta, t, f(t)f(t)f, left parenthesis, t, right parenthesis increases by \dfrac{\Delta t}{10.5}

10.5

Δt

​

start fraction, delta, t, divided by, 10, point, 5, end fraction.

Therefore, for f(t)f(t)f, left parenthesis, t, right parenthesis to increase by 111, we need \Delta t=10.5Δt=10.5delta, t, equals, 10, point, 5. In other words, the ttt-interval we are looking for is 10.510.510, point, 5 minutes.

Hint #33 / 3

Summary

The number of people who yawned quadruples every 10.510.510, point, 5 minutes.

7 0
3 years ago
Can someone help me with this please
makvit [3.9K]
Rise over run
mx+b
1/3x-1
6 0
3 years ago
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