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2 years ago

A TRAIN TRAVELING AT A CONSTANT RATE OF SPEED COVERS 360 KILOMETERS IN 6 HOURS WHAT IS THE SPEED OF THE TRAIN

Mathematics

2 answers:

never [62]2 years ago

8 0

Answer:

60kmh

Step-by-step explanation:

in order to answer this, we do some simple division.

360 divided by 6 is 60, so the train is going 60km an hour.

FromTheMoon [43]2 years ago

5 0

Answer:

60 km/hr

Step-by-step explanation:

Speed = distance / time

Speed = 360 km / 6 hr

Speed = 60 km/hr

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Answer:

If this is a proof then here is the answer.

Angle ABD is Congruent to Angle CBD = Given

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Line Segment BD is Congruent to Line Segment BD = Reflexive Property

Line Segment AB is Congruent to Linge Segment CB = Angle-Side-Angle or ASA

Step-by-step explanation:

Lucky for you, I just learned this also ;)

Since you are given your first two directions, put them down as GIVEN in the proof.

Next, Since ABD and CBD are congruent angles, you can assume that it is an angle bisector since angle bisectors always bisect equally.

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2 years ago

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A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per

Tems11 [23]

Answer:

The pressure is changing at $\frac{dP}{dt}=3.68$

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ and we want to find at what rate is the pressure changing.

The equation that model this situation is

$PV^{1.4}=k$

Differentiate both sides with respect to time t.

$\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\$

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

$\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)$

Apply this rule to our expression we get

$V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0$

Solve for $\frac{dP}{dt}$

$V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}$

when P = 23 kg/cm2, V = 35 cm3, and $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ this becomes

$\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68$

The pressure is changing at $\frac{dP}{dt}=3.68$ .

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Answer:

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2 years ago

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