Is 4.1 greater than 4

Mathematics

2 answers:

EleoNora [17]3 years ago

6 0

Answer: yes it is.

let me know if you need anything else

Step-by-step explanation:

labwork [276]3 years ago

6 0

Answer:

<em>I believe so soldier~!</em>

Step-by-step explanation:

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The radius of one sphere is twice as great as the radius of a second sphere. Find the ratio of their volumes. 1/2 1/4 1/8 1/16

Anton [14]

Let's pick some simple points with which to set up an example for ourselves for this. Let's let the smaller radius be 1, and the larger, twice that, be 2. The radius itself is a single unit measure; in other words, it's measured as inches, feet, cm, etc., while the volume is a cubed measure. Volume is measured in inches cubed, feet cubed, cm cubed, etc. Therefore, if we have the radii measuring 1:2, we simply cube those single unit measures to find the ratio of their volumes. 1 cubed is 1, and 2 cubed is 8. So your answer for this is 1/8.

5 0

4 years ago

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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 $\frac{dP}{dt}$

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