Find the value of the given variable (-4, r) (-8, 3) m=5

Just question 4 please. I NEED TO SHOW MY WORK AS TO HOW I GOT THE ANSWER!!!! Stuck someone please help?!!

Galina-37 [17]

Well first you wanna see how many pints are in 3 quarts, which is 6 because there are 2 pints in a quart.

3 0

3 years ago

John is 3 years older than Jim. Jim is 4 years less than half of Dana’s age. How old is each person if their ages add up to 167?

kogti [31]

Solution:

we are given that

John is 3 years older than Jim. Jim is 4 years less than half of Dana’s age.

Let the age of Jim be x years then we can say

Age of John $=x+3$

If diana Age is D then

$x=D/2-4$

It can be re-wriiten as

Diana age $D=2x+8$

Given that their ages add up to 167

$x+x+3+2x+8=167\\
\\
4x+11=167\\
\\
4x=167-11\\
\\
4x=156\\
\\
x=39\\$

Hence jim is 39 nyears old, John is 39+3=42 years old. and Diana is 2*39+8=78+8=86 years old.

3 0

2 years ago

HELPPP ASAP!!! GIVING BRAINLIEST!!!!

diamong [38]

Answer:

x = -4 and x = -2

Step-by-step explanation:

5 0

2 years ago

How many solutions does the system of equations shown below have?

rjkz [21]

Answer:

one solution with a y value of 5

Step-by-step explanation:

/| x+y-4 = 0| x-y-6 = 0

We try to solve the equation: x+y-4 = 0

x+y-4 = 0 // - x-4

y = -(x-4)

y = 4-x

We insert the solution into one of the initial equations of our system of equations

We get a system of equations:

/| x+x-6-4 = 0| y = 4-x

2*x-10 = 0 // + 10

2*x = 10 // : 2

x = 10/2

x = 5

We insert the solution into one of the initial equations of our system of equations

For y = 4-x:

y = 4-5

y = -1

We get a system of equations:

/| y = -1| x = 5

4 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