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Anna11 [10]
3 years ago
5

A school is planning a field trip for 142 people. The trip will use 6 drivers and two types of vehicles:buses and vans. A bus ca

n seat 51 passengers. A van can seat 10 passengers. Write and solve a system of equations to find how many buses and vans will be needed
Mathematics
2 answers:
Fynjy0 [20]3 years ago
5 0
So 51+51=102 and then 10+10+10+10=40, and 102+40=142. So they will need to take 2 buses and 4 vans. 
                                             hope I helped
elixir [45]3 years ago
3 0

Answer:  The answer is 2 buses and 4 vans.


Step-by-step explanation:  Let 'b' be the number of buses and 'v' be the number of vans that will be needed.

The trip will use 6 drivers, and we need one driver for each vehicle, so number of drivers is equal to the number of vehicles used.i.e.,

b+v=6.

Now, number of passengers in buses is 51b and number of passengers in vans is 10v. Since there are total 142 people, so

51b+10v=142.

Substituting the value of 'v' from forst equation in the second equation, we have

51b+10(6-b)=142\\\\\Rightarrow 51b+60-10b=142\\\\\Rightarrow 41b=82\\\\\Rightarrow b=2.

Again, from the first equation, we get

v=6-2=4.

Thus, there are 2 buses and 4 vans in the trip.


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an inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a
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the rate of change of the water depth when the water depth is 10 ft is;  \mathbf{\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi} \  \ ft/s}

Step-by-step explanation:

Given that:

the inverted conical water tank with a height of 20 ft and a radius of 8 ft  is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.

We are meant to find the  rate of change of the water depth when the water depth is 10 ft.

The diagrammatic expression below clearly interprets the question.

From the image below, assuming h = the depth of the tank at  a time t and r = radius of the cone shaped at a time t

Then the similar triangles  ΔOCD and ΔOAB is as follows:

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\dfrac{h}{r}= \dfrac{5}{2}

\dfrac{h}{r}= 2.5

h = 2.5r

r = \dfrac{h}{2.5}

The volume of the water in the tank is represented by the equation:

V = \dfrac{1}{3} \pi r^2 h

V = \dfrac{1}{3} \pi (\dfrac{h^2}{6.25}) h

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The rate of change of the water depth  is :

\dfrac{dv}{dt}= \dfrac{\pi r^2}{6.25}\  \dfrac{dh}{dt}

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\dfrac{dv}{dt}= - 4  \ ft^3/sec

Therefore,

-4 = \dfrac{\pi r^2}{6.25}\  \dfrac{dh}{dt}

the rate of change of the water at depth h = 10 ft is:

-4 = \dfrac{ 100 \ \pi }{6.25}\  \dfrac{dh}{dt}

100 \pi \dfrac{dh}{dt}  = -4 \times 6.25

100  \pi \dfrac{dh}{dt}  = -25

\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi}

Thus, the rate of change of the water depth when the water depth is 10 ft is;  \mathtt{\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi} \  \ ft/s}

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