What is the slope of a line that passes through the points (-3, 2) and (-6, 5)?

6 less than one eighth of y is x

faltersainse [42]

Answer:

1/8y-6=x

Step-by-step explanation:

6 0

2 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 $\frac{dP}{dt}$

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

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 $\frac{dP}{dt}=3.68$ .

7 0

3 years ago

In triangle abc, the angle c is six times as large as angle

mote1985 [20]

The interior angles of a triangle add up to 180 degrees

a = 1st angle, b = 2nd angle, c = 3rd angle

c = 6a
b = a + 60
a + b + c = 180

a + (a + 60) + (6a) = 180
8a + 60 = 180
8a = 180 - 60
8a = 120
a = 120/8
a = 15 <=== here is one angle

b = a + 60
b = 15 + 60
b = 75 <=== and another one

c = 6a
c = 6(15)
c = 90 <=== and the last angle

3 0

2 years ago

Solve 5x^2=10x + 4<br><br> Help please I literally forgot what topic this is! Thanksssss !!!!

faltersainse [42]

X = 4/15. The topic is solving for unknowns in an equation.

5 0

3 years ago

What does air smell like?

dmitriy555 [2]

Nothing except when it has been polluted

8 0

2 years ago

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