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3 years ago

PLEASE MAKE IT QUICK! HELP ME PLZZZ! ANSWER WITH SOLUTION

Mathematics

2 answers:

Serga [27]3 years ago

7 0

Answer:

is 24

Step-by-step explanation:

it don't have cero

balu736 [363]3 years ago

4 0

1 galon = 8 pints

30 galon = 240 pints

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quester [9]

To find Angle A we use cosine

cos ∅ = adjacent / hypotenuse

From the question

The adjacent is 17

The hypotenuse is 38

So we have

cos A = 17/38

A = cos-¹ 17/38

A = 63.4

<h3>A = 63° to the nearest degree</h3>

To find Angle C we use sine

sin ∅ = opposite / hypotenuse

From the question

The opposite is 17

The hypotenuse is 38

So we have

sin C = 17/38

C = sin-¹ 17/38

C = 26.57

<h3>C = 27° to the nearest degree</h3>

Hope this helps you

7 0

3 years ago

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 $\frac{dP}{dt}$

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