All categories

3 years ago

Write your own situation for the expression c = 9g

1 answer:

4 0

Ivy is going to the fair with a friend. They will buy cotton candy. They buy 9 bags of it. Each bag cost 2 dollars . What is C the total cost of all the cotton candy if G=2?

You might be interested in

A pack of folders has a length of 5 inches, a width of 12 inches, and a height of 1 inch. The pack of folders will be shipped in

777dan777 [17]

Answer:

A) Each pack of folders has a volume of 60 cubic inches.

B) The box has a volume of about 720 cubic inches

D) If the box help 20 packs of folders, it would have a volume of about 1,200 cubic inches.

Step-by-step explanation:

Verify each statement

A) Each pack of folders has a volume of 60 cubic inches.

The statement is True

Because

The volume of each pack of folders is equal to

$V=(5)(12)(1)=60\ in^{3}$

B) The box has a volume of about 720 cubic inches

The statement is True

Because

The volume of the box is equal to the volume of one pack of folders multiplied by 12

$V=(12)60=720\ in^{3}$

C) If the box held 15 packs of folders, it would have a volume of about 1,200 cubic inches

The statement is False

Because

Applying proportion

$\frac{12}{720}\frac{packs}{in^{3}}=\frac{15}{x}\frac{packs}{in^{3}}\\ \\x=720*15/12\\ \\x=900\ in^{3}$

$900\ in^{3}\neq 1,200\ in^{3}$

D) If the box help 20 packs of folders, it would have a volume of about 1,200 cubic inches.

The statement is True

Because

Applying proportion

$\frac{12}{720}\frac{packs}{in^{3}}=\frac{20}{x}\frac{packs}{in^{3}}\\ \\x=720*20/12\\ \\x=1,200\ in^{3}$

$1,200\ in^{3}= 1,200\ in^{3}$

E) Each pack of folders has a volume of 24 cubic inches.

The statement is False

Because

The volume of each pack of folders is equal to

$V=(5)(12)(1)=60\ in^{3}$

7 0

3 years ago

Given the parent function f(x)=|x|, write the equation of the function g(x) whose graph is pictured

kenny6666 [7]

By analyzing the graph of the transformed function, we can see that:

g(x) = 2*|x - 2| - 4

<h3>How to get the function g(x)?</h3>

In the image we have two graphs, the yellow one is the graph of g(x).

First, analyzing the vertex we can see the translation used, you need to remember:

Horizontal translation:

For a general function f(x), a horizontal translation of N units is written as:

g(x) = f(x + N).

If N is positive, the shift is to the left.
If N is negative, the shift is to the right.

Vertical translation:

For a general function f(x), a vertical translation of N units is written as:

g(x) = f(x) + N.

If N is positive, the shift is upwards.
If N is negative, the shift is downwards.

We can see that the new vertex is at (2, -4), so the translation is 2 units to the right and 4 units down, then we have:

g(x) = |x - 2| - 4

Now, you also can see that the yellow function is narrower, such that for each increase in the x-unit, we have an increase of 2 in the y-axis, then we need to multiply by 2 the absolute value part:

g(x) = 2*|x - 2| - 4

This is the transformed function.

If you want to learn more about transformations, you can read:

brainly.com/question/4289712

5 0

2 years ago

A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in the solution. Water containing1 lb

devlian [24]

Answer:

(a) The amount of salt in the tank at any time prior to the instant when the solution begins to overflow is $\left(1-\frac{4000000}{\left(200+t\right)^3}\right)\left(200+t\right)$ .

(b) The concentration (in lbs per gallon) when it is at the point of overflowing is $\frac{121}{125}\:\frac{lb}{gal}$ .

(c) The theoretical limiting concentration if the tank has infinite capacity is $1\:\frac{lb}{gal}$ .

Step-by-step explanation:

This is a mixing problem. In these problems we will start with a substance that is dissolved in a liquid. Liquid will be entering and leaving a holding tank. The liquid entering the tank may or may not contain more of the substance dissolved in it. Liquid leaving the tank will of course contain the substance dissolved in it. If Q(t) gives the amount of the substance dissolved in the liquid in the tank at any time t we want to develop a differential equation that, when solved, will give us an expression for Q(t).

The main equation that we’ll be using to model this situation is:

Rate of change of Q(t) = Rate at which Q(t) enters the tank – Rate at which Q(t) exits the tank

where,

Rate at which Q(t) enters the tank = (flow rate of liquid entering) x (concentration of substance in liquid entering)

Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x (concentration of substance in liquid exiting)

Let $C$ be the concentration of salt water solution in the tank (in $\frac{lb}{gal}$ ) and $t$ the time (in minutes).

Since the solution being pumped in has concentration $1 \:\frac{lb}{gal}$ and it is being pumped in at a rate of $3 \:\frac{gal}{min}$ , this tells us that the rate of the salt entering the tank is

$1 \:\frac{lb}{gal} \cdot 3 \:\frac{gal}{min}=3\:\frac{lb}{min}$

But this describes the amount of salt entering the system. We need the concentration. To get this, we need to divide the amount of salt entering the tank by the volume of water already in the tank.

$V(t)$ is the volume of brine in the tank at time t. To find it we know that at t = 0 there were 200 gallons, 3 gallons are added and 2 are drained, and the net increase is 1 gallons per second. So,

$V(t)=200+t$

Therefore,

The rate at which $C(t)$ enters the tank is

$\frac{3}{200+t}$

The rate of the amount of salt leaving the tank is

$C\:\frac{lb}{gal} \cdot 2 \:\frac{gal}{min}+C\:\frac{lb}{gal} \cdot 1\:\frac{gal}{min}=3C\:\frac{lb}{min}$

and the rate at which $C(t)$ exits the tank is

$\frac{3C}{200+t}$

Plugging this information in the main equation, our differential equation model is:

$\frac{dC}{dt} =\frac{3}{200+t}-\frac{3C}{200+t}$

Since we are told that the tank starts out with 200 gal of solution, containing 100 lb of salt, the initial concentration is

$\frac{100 \:lb}{200 \:gal} =0.5\frac{\:lb}{\:gal}$

Next, we solve the initial value problem

$\frac{dC}{dt} =\frac{3-3C}{200+t}, \quad C(0)=\frac{1}{2}$

$\frac{dC}{dt} =\frac{3-3C}{200+t}\\\\\frac{dC}{3-3C} =\frac{dt}{200+t} \\\\\int \frac{dC}{3-3C} =\int\frac{dt}{200+t} \\\\-\frac{1}{3}\ln \left|3-3C\right|=\ln \left|200+t\right|+D\\\\$

We solve for C(t)

$C(t)=1+D(200+t)^{-3}$

D is the constant of integration, to find it we use the initial condition $C(0)=\frac{1}{2}$

$C(0)=1+D(200+0)^{-3}\\\frac{1}{2} =1+D(200+0)^{-3}\\D=-4000000$

So the concentration of the solution in the tank at any time t (before the tank overflows) is

$C(t)=1-4000000(200+t)^{-3}$

(a) The amount of salt in the tank at any time prior to the instant when the solution begins to overflow is just the concentration of the solution times its volume

$(1-4000000(200+t)^{-3})(200+t)\\\left(1-\frac{4000000}{\left(200+t\right)^3}\right)\left(200+t\right)$

(b) Since the tank can hold 500 gallons, it will begin to overflow when the volume is exactly 500 gal. We noticed before that the volume of the solution at time t is $V(t)=200+t$ . Solving the equation

$200+t=500\\t=300$

tells us that the tank will begin to overflow at 300 minutes. Thus the concentration at that time is

$C(300)=1-4000000(200+300)^{-3}\\\\C(300)= \frac{121}{125}\:\frac{lb}{gal}$

(c) If the tank had infinite capacity the concentration would then converge to,

$\lim_{t \to \infty} C(t)= \lim_{t \to \infty} 1-4000000\left(200+t\right)^{-3}\\\\\lim _{t\to \infty \:}\left(1\right)-\lim _{t\to \infty \:}\left(4000000\left(200+t\right)^{-3}\right)\\\\1-0\\\\1$