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Travka [436]
3 years ago
9

The deck of a container ship is 1,200 feet long and 300 feet wide. If one container is 30 feet long and 10 feet wide, and contai

ners can be stacked four high, how many containers can the ship carry at a time
Mathematics
1 answer:
Gemiola [76]3 years ago
3 0

Answer: 4800 containers

Step-by-step explanation:

We can see that along the length, the ship can carry (1200 / 30) = 40 containers and along the width, it can carry (300/10) = 30 containers.

Since the containers can be stacked four high, the number of containers that the ship can carry at a time would be:

= 40 × 30 × 4.

= 4800 containers

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Help me fast asap this is due by the end of class
Rufina [12.5K]

Answer:

D

Step-by-step explanation:

-12/6= -2

x^10/x^8= x^2

y^3/y= y^2

5 0
2 years ago
A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k
Scrat [10]

Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

a.

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

5 0
3 years ago
4 students equally share 2/3 of a pizza
oee [108]

\frac{2}{3} *4

=\frac{2*4}{3}

=\frac{8}{3}

2.67 or 2 \frac{2}{3}

6 0
3 years ago
A piece of wood is 12 feet long and a nail is 3 inches long. The piece of wood is how many times as long as the nail?
Softa [21]
The piece of wood is 48 times as long as the nail (12 feet = 144 inches) because 3 goes into 144 48 times.
7 0
3 years ago
Read 2 more answers
If DF=78, DE=5x-9, and EF=2x+10, find DE.
Kazeer [188]

Given that E is a point between Point D and F, the numerical value of segment DE is 46.

<h3>What is the numerical value of DE?</h3>

Given the data in the question;

  • E is a point between point D and F.
  • Segment DF = 78
  • Segment DE = 5x - 9
  • Segment EF = 2x + 10
  • Numerical value of DE = ?

Since E is a point between point D and F.

Segment DF = Segment DE + Segment EF

78 = 5x - 9 + 2x + 10

78 = 7x + 1

7x = 78 - 1

7x = 77

x = 77/7

x = 11

Hence,

Segment DE = 5x - 9

Segment DE = 5(11) - 9

Segment DE = 55 - 9

Segment DE = 46

Given that E is a point between Point D and F, the numerical value of segment DE is 46.

Learn more about equations here: brainly.com/question/14686792

#SPJ1

3 0
1 year ago
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