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10 months ago

The scale factor from S to M is 3/2 What is the scale factor from M to S

Mathematics

1 answer:

natita [175]10 months ago

3 0

we conclude that if the scale factor from S to M is 3/2, then the scale factor from M to S is 2/4.

<h3></h3><h3>What is the scale factor from M to S?</h3>

Suppose we have a figure S. If we apply a stretch of scale factor K to our figure S, we can say that all the dimensions of figure S are multiplied by K.

So, if S represents the length of a bar, then after the stretch we will get a bar of length M, such that:

M = S*K

If that scale factor is 3/2, then we have the case of the problem:

M = (3/2)*S

We can isolate S in the above relation:

(2/3)*M = S

Now we have an equation (similar to the first one) that says that the scale factor from M to S is 2/3.

Then we conclude that if the scale factor from S to M is 3/2, then the scale factor from M to S is 2/4.

If you want to learn more about scale factors:

brainly.com/question/25722260

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Answer:

b. (3,0)

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Are solutions to the system of inequalities

Step-by-step explanation:

b. (3,0)

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e. (5,-6)

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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 $\frac{dP}{dt}$

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