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3 years ago

To start, which measurement does Tommy need to

Mathematics

2 answers:

Luba_88 [7]3 years ago

8 0

Answer:

C. volume

Step-by-step explanation:

when youre starting your measurements you want to start the volume first.

sesenic [268]3 years ago

6 0

Answer:

Step-by-step explanation:

got it right

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If f(x) = 3 ^ x - 7 and g(x) = x ^ 2 + 11 , the value of f(2) - g(7) is:

Maurinko [17]

Answer:

The value of $f(2) - g(7)=-58$

Step-by-step explanation:

We have

$f(x) = 3 ^ x - 7\\\\f(2) = 3 ^ 2 - 7\\\\=9-7\\\\=2\\\\g(x) = x ^ 2 + 11\\\\g(7) = 7 ^ 2 + 11\\\\=49+11\\\\=60\\\\f(2) - g(7)=2-60\\\\=-58$

4 0

3 years ago

Please help!!! will give brainliest

mina [271]

Answer:

The answer is B

Step-by-step explanation:

3 to the forth power = 81

3 to the second power = 9

81/9 = 9

8 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 $\frac{dP}{dt}$

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