15. Find the following bases.

Solve for m. <br> 2(5m + 4) = 2 (3m - 10)

agasfer [191]

2(5m+4)=2(3m-10)
10m+8= 6m-20
-6m -6m
4m+8= -20
-8 -8
4m= -28
m = -7

3 0

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 $\frac{dP}{dt}$

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

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

3 years ago

Which equation represents a population of 210 animals that decreases at an annual rate of 14%

soldi70 [24.7K]

Answer:

The equation i.e. used to denote the population after x years is:

P(x) = 490(1 + 0.200 to the power of x

Step-by-step explanation:

This problem could be modeled with the help of a exponential function.

The exponential function is given by:

P(x) = ab to the power of x

where a is the initial value.

and b=1+r where r is the rate of increase or decrease.

Here the initial population of the animals are given by: 490

i.e. a=490

Also, the rate of increase is: 20%

i.e. r=20%

i.e. r=0.20

Hence, the population function i.e. the population of the animals after x years is:

P(x) = 490(1 + 0.200 to the power of x

6 0

2 years ago

For dessert you get to choose between 3 types of pie, 2 types of ice cream, and 3 drinks. How many pie, ice cream, and drink com

sladkih [1.3K]

You simply multiply all the options
3x2x3= 18 possibilities

4 0

3 years ago

Read 2 more answers

The two-way table shows information about the subjects studied

son4ous [18]

Answer:

a) 62

b) 31/45

Step-by-step explanation:

a) The table tells you there are 42 males, and 20 more females that study biology.

42 +20 = 62 . . . are male or study biology or both

__

b) Of the total of 90 people, 62 are male or study biology (or both). The probability that any person is in that category is ...

62/90 = 31/45 = P(male or biology)

4 0

2 years ago