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3 years ago

I need help with 1 and 2

Mathematics

1 answer:

Eddi Din [679]3 years ago

7 0

1. X take away 12. You're subtracting in this problem, while you add in all the other problems. The expression would be x-12.

2. The coefficient represents the cost to print the photos. It costs 0.25 to print "p" amount of photos.

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Andrea's dad ran 12 miles and 75 minutes on a treadmill if he runs at that rate how far could he run in 30 minutes

Alexus [3.1K]

Answer: 187.5

Step-by-step explanation: Divide 75 by 12 to get 6.25 then multiply that by 30 to get 187.5.

6 0

3 years ago

Find the value of 7w-2 given that - 2w-4=6 .

Vesnalui [34]

Answer:

w= -5

Step-by-step explanation:

-2w - 4 = 6

-2w - 4 + 4 = 6 + 4
-2w = 10

Isolate "w":

$\frac{-2w}{-2} = \frac{10}{-2}$

w = -5

4 0

3 years ago

Find an equation of variation in which y varies jointly as x and z and inversely as the product of w and p, where yequalsstartf

leva [86]

To solve this problem you must apply the proccedure shown below:

1. You have that $y$ varies jointly as $x$ and $z$ and inversely as the product of $w$ and $p$ . Therefore, you can write the following equation, where $k$ is the constant of proportionality:

$y=k(\frac{xz}{wp} )$

2. Now, you must solve for the constant of proportionality, as following:

$k=\frac{ywp}{xz}$

3. Susbtiute values:

$y=\frac{7}{28} \\ x=7\\ z=4\\ w=7\\ p=8$

$k=\frac{(\frac{7}{28})(7)(8))}{(7)(4)} =0.5$

4. Substitute the value of the constant of proportionality into the equation:

$y=0.5(\frac{xz}{wp})$

The answer is: $y=0.5(\frac{xz}{wp})$

7 0

3 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

4 years ago

~I ~ need ~ help ~ once ~ again~

ruslelena [56]

A. 12pi meters.

Circumference = pi x diameter (pi x 12 = 12pi)
Diameter = 2 x radius (2 x 6 = 12)

8 0

3 years ago