7/3x+1/3x=1+5/3x<br>plz help

What is the relationship between area in the scale drawing and area on the on the actual tennis court if the scale is 1in.: 5ft.

sveticcg [70]

Given:

Length of the scale = 15.6 in.

Width of the scale = 7.2 in.

Scale of drawing = 1 in. : 5ft.

To find:

The ratio of area of the actual court to the area of the drawing (as a unit rate).

And to check whether it is the same as the ratio of length of the actual court to the length of the drawing.

Step-by-step explanation:

We have,

1 in. = 5ft.

Now, using this scale we get

15.6 in. = (15.6 × 5) ft =78 ft.

7.2 in. = (7.2 × 5) ft = 36 ft.

So, the actual length and width of tennis court are 78 ft and 36 ft respectively.

Area of actual tennis court is

$Area=length\times width$

$Area=78\times 36$

$Area=2808\text{ ft}^2$

The area of drawing is

$Area=15.6\times 7.2$

$Area=112.32\text{ in.}^2$

Now, ratio of area of the actual court to the area of the drawing (as a unit rate) is

$\dfrac{2808}{112.32}=\dfrac{25}{1}=25:1$

Ratio of area is 25:1 and ratio of length is 5:1 both area not same.

Therefore, the ratio of area of the actual court to the area of the drawing (as a unit rate) is 25:1.

7 0

3 years ago

What is 5 3/8 - 3 2/5 ( like fractions w a whole number )

Leona [35]

Answer:

1 39/40

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Explain how to find the greatest number of identical arrangements that can be made with 54 roses and 42 tulips with no flowers l

Basile [38]

Work out the Greatest Common Factor for both numbers

54 421×54 1×422×27 2×213×18 3×146×9 6×7

The common factors are: 1, 2, 3 and 6
The greatest common factor is 6
There are 6 identical arrangements that can be made

There will be 9 roses and 7 tulips in each arrangement

3 0

3 years ago

Read 2 more answers

When two numbers are added together,their sum is 6750.The difference between the two number is 250.Find the larger number.

dimaraw [331]

Bigger number is 3625 and smaller number is 3125

8 0

3 years ago

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows

4 0

3 years ago

7/3x+1/3x=1+5/3xplz help ​

7/3x+1/3x=1+5/3xplz help