Plz math help me in class !!!!!

3) When Jackie took her children to the movie theater, she bought one box of candy and three small drinks for

Nezavi [6.7K]

Answer:

c + 3d = 14.75

2c + 5d = 26

A box of candy is $4.25 and a drink is $3.50

Step-by-step explanation:

Let c represent the cost of a box of candy and let d represent the cost of a drink

c + 3d = 14.75

2c + 5d = 26

Solve by elimination by multiplying the top equation by -2

-2c - 6d = -29.5

2c + 5d = 26

Add them together

-d = -3.5

d = 3.5

Plug in 3.5 as d into one of the equations to solve for c

c + 3d = 14.75

c + 3(3.5) = 14.75

c + 10.5 = 14.75

c = 4.25

So, a box of candy is $4.25 and a drink is $3.50

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 $\frac{dP}{dt}$

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