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2 years ago

The price of a car was decreased by 16% to £2400. What was the price before the decrease?

Mathematics

1 answer:

miss Akunina [59]2 years ago

3 0

Answer:

£2857.14

Step-by-step explanation:

The price after the decrease is 84% of the original price, so we have ...

£2400 = 0.84 · (original price)

£2400/0.84 = original price ≈ £2857.14

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Vanna is saving for a trip. The hotel room will be $298.17 for 3 nights, and there will be additional fees. What is her daily co

docker41 [41]

Answer:

Step-by-step explanation:

Total cost for the three nights

Total_3 = $298.17 + 3*u

Where u represents the unknown fees for a single day

To find the daily cost, we divide the previous equation by three

Daily cost = ($298.17 + 3*u)/3

Daily cost = ($99.39 + u)

So, if we create an inequality for the daily cost

Let x = Daily cost

x > $99.39

She will pay more than $99.39 per night

7 0

2 years ago

What is the coefficient of each monomial? a. 5pk b. f c. –9t d. –j

Semenov [28]

Step-by-step explanation:

Look at the picture

a. 5pk → 5

b. f = 1f → 1

c. -9t → -9

d. -j = -1j → -1

5 0

2 years ago

Read 2 more answers

The diameter of Circle Q terminates on the circumference of the circle at (0,3) and (0,-4). Write the equation of the circle in

Gnesinka [82]

First, determine the center of the circle by getting the midpoint of the points given for the circumference.
midpoint = ((0 + 0)/2, (3 + -4)/2)
midpoint (0, -0.5)
Then, we get the radius by determining the distance from either of the circumferential point to the center.
radius = √(0 - 0)² + (3 +4)² = 7
The equation for the circle would be,
x² + (y + 0.5)² = 7²

8 0

2 years ago

A 400 gallon tank initially contains 100 gal of brine containing 50 pounds of salt. Brine containing 1 pound of salt per gallon

posledela

Answer:

The amount of salt in the tank when it is full of brine is 393.75 pounds.

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 y(t) be the amount of salt (in pounds) in the tank at time t (in seconds). Then we can represent the situation with the below picture.

Then the differential equation we’re after is

$\frac{dy}{dt} = (Rate \:in)- (Rate \:out)\\\\\frac{dy}{dt} = 5 \:\frac{gal}{s} \cdot 1 \:\frac{pound}{gal}-3 \:\frac{gal}{s}\cdot \frac{y(t)}{V(t)} \:\frac{pound}{gal}\\\\\frac{dy}{dt} =5\:\frac{pound}{s}-3 \frac{y(t)}{V(t)} \:\frac{pound}{s}$

$V(t)$ is the volume of brine in the tank at time t. To find it we know that at time 0 there were 100 gallons, 5 gallons are added and 3 are drained, and the net increase is 2 gallons per second. So,

$V(t)=100 + 2t$

We can then write the initial value problem:

$\frac{dy}{dt} =5-\frac{3y}{100+2t} , \quad y(0)=50$

We have a linear differential equation. A first-order linear differential equation is one that can be put into the form

$\frac{dy}{dx}+P(x)y =Q(x)$

where P and Q are continuous functions on a given interval.

In our case, we have that

$\frac{dy}{dt}+\frac{3y}{100+2t} =5 , \quad y(0)=50$

The solution process for a first order linear differential equation is as follows.

Step 1: Find the integrating factor, $\mu \left( x \right)$ , using $\mu \left( x \right) = \,{{\bf{e}}^{\int{{P\left( x \right)\,dx}}}$

$\mu \left( t \right) = \,{{e}}^{\int{{\frac{3}{100+2t}\,dt}}}\\\int \frac{3}{100+2t}dt=\frac{3}{2}\ln \left|100+2t\right|\\\\\mu \left( t \right) =e^{\frac{3}{2}\ln \left|100+2t\right|}\\\\\mu \left( t \right) =(100+2t)^{\frac{3}{2}$

Step 2: Multiply everything in the differential equation by $\mu \left( x \right)$ and verify that the left side becomes the product rule $\left( {\mu \left( t \right)y\left( t \right)} \right)'$ and write it as such.

$\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+\frac{3y}{100+2t}\cdot \left(100+2t\right)^{\frac{3}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+3y\cdot \left(100+2t\right)^{\frac{1}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})=5\left(100+2t\right)^{\frac{3}{2}}$

Step 3: Integrate both sides.

$\int \frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})dt=\int 5\left(100+2t\right)^{\frac{3}{2}}dt\\\\y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }+ C$

Step 4: Find the value of the constant and solve for the solution y(t).

$50 \left(100+2(0)\right)^{\frac{3}{2}}=(100+2(0))^{\frac{5}{2} }+ C\\\\100000+C=50000\\\\C=-50000$

$y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }-50000\\\\y(t)=100+2t-\frac{50000}{\left(100+2t\right)^{\frac{3}{2}}}$

Now, the tank is full of brine when:

$V(t) = 400\\100+2t=400\\t=150$

The amount of salt in the tank when it is full of brine is

$y(150)=100+2(150)-\frac{50000}{\left(100+2(150)\right)^{\frac{3}{2}}}\\\\y(150)=393.75$

6 0

2 years ago

In a survey of 2,300 people who owned a certain type of car, 1,610 said they would buy that type of car again. What percent of t

trapecia [35]

Answer:

In a survey of 2,300 people who owned a certain type of car, 1,610 said they would buy that type of car again. What percent of the people surveyed were satisfied with the car?

1610/2300x100=

70%

Step-by-step explanation:

total number of people for the survey= 2300

those that wanted to buy the car= 1610

percentage= 1610/2300x 100=0.7x 100

percentage= 70%

3 0

2 years ago