All categories

3 years ago

Please help and u crush will marry u!!!!!!!

Mathematics

1 answer:

Vedmedyk [2.9K]3 years ago

5 0

weii ni yo entiendo esa mamada:u

You might be interested in

What is the polynomial of 5x3-4x2+7x

olga nikolaevna [1]

Answer:

Step-by-step explanation:

7+7x

5 0

3 years ago

-1(n2 + 3) if n = -3.

Soloha48 [4]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

-4(3+7m) please help i will give brainlest and do not answer just to get points this is DUE IN FIVE MINUTES

lianna [129]

Answer:

-28m-12

Step-by-step explanation:hope that helps :)

=(−4)(3+7m)

=(−4)(3)+(−4)(7m)

=−12−28m

=−28m−12

6 0

3 years ago

Read 2 more answers

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 $\frac{dP}{dt}$

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