Rewrite the quadratic function below in standard form y=3(x+2)(x-3)

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

4 years ago

The quotient of 56 and n is multiplied by 3.

Nutka1998 [239]

Answer:

Equivalent expression using n to represent the unknown number is: $\mathbf{(\frac{56}{n})\times 3\\}$

Step-by-step explanation:

The quotient of 56 and n is multiplied by 3.

Create an equivalent expression using n to represent the unknown number.

We need to write equivalent expression of The quotient of 56 and n is multiplied by 3.

We will do this step by step

First we will find:

The quotient of 56 and n

It can be written in mathematical form is: $\frac{56}{n}\\$

Now, we will find

The quotient of 56 and n is multiplied by 3.

It can be written in mathematical form as: $(\frac{56}{n})\times 3\\$

So, equivalent expression using n to represent the unknown number is: $\mathbf{(\frac{56}{n})\times 3\\}$

5 0

3 years ago

A pond holds 1920 gallons of water. There is a leak and the amount of water drops 20 gallons every hour. How much water is in th

Ivahew [28]

Answer:

20 times an hour so 20 times 48 is 960 1920-960 is 960 gallons so there is 960 gallons of water left

7 0

3 years ago

Read 2 more answers

PLEASE HELPP MEE!!!! MATH!!! PLEASEEE HELP ME!!!!!

IRISSAK [1]

Answer:

The solution is (3,2)

Step-by-step explanation:

x - y = 1 →equation 1

-x + 3y = 3 →equation 2

by elimination method,

x - y = 1

-x + 3y = 3

-----------------

0 + 2y = 4

2y = 4

y = 4/2

y = 2

substitute y=2 in equation 1

x - y = 1

x - 2 = 1

x = 1 + 2

x = 3

∴(3,2) is the solution

5 0

3 years ago

A regular hexagon is inscribed in a circle and another regular hexagon is circumscribed about the same circle. What is the ratio

Anna007 [38]

4/3 of the area of the larger hexagon to the area of the smaller hexagon

Let the sides of the interior hexagon be 2 cm long, then this hexagon is made up of 6 equilateral triangles of side = 2.

The area of this hexagon = 6 * 1 * sqrt3 = 6 sqrt3 ( as the triangle is made up of 2 60-30-90 triangles)

The exterior hexagon is made up of 6 equilateral triangles with altitude 2 and the area = 6 * 2 * (2 / sqrt3 ) = 24 / sqrt3

area exterior hex / interior hex = 24 / sqrt3 * 1 / 6sqrt3 = 24/18

= 4/3

Required ratio = 4/3 answer

Learn more about Equations:

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3 0

2 years ago