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Nimfa-mama [501]
3 years ago
5

A local park has a grassy playground that is 3/5 mile long and 1/3 mile wide. What part of a square mile does the playground cov

er? Enter your answer, in simplest form, in the boxes.
?

----mi²

?
Mathematics
1 answer:
adell [148]3 years ago
5 0

Answer:

The playground covers \frac{1}{5} of a square mile


Step-by-step explanation:

The playground is a rectangular ground. The area of a rectangle is given by legth * width, hence,

The area of the playground is length * width

Area = \frac{3}{5}*\frac{1}{3}=\frac{3}{15}=\frac{1}{5} sq. mi


Hence the playground cover \frac{1}{5} of a square mile

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Find the number that belongs
tangare [24]

Answer:

2.7

Step-by-step explanation:

322/7=46

46*125=2.71739

round off

2.7

4 0
3 years ago
The ratio of two numbers is 4 to 3. The sum of the numbers is 49. What are the two numbers?​
Irina-Kira [14]

Answer:

28/21

Step-by-step explanation:

28/21 is the equivalent to 4/3.

We know this is the fraction we are looking for because 28 + 21 = 49

Hope that helps!

4 0
2 years ago
A basketball court has a rectangular shape with a length of 94 feet and a width of 50 feet. Janet made two scale models if the r
Irina-Kira [14]

Answer:

Step-by-step explanation:

Note that there are two scale models with each of ratio of 1/2 and 1/16 respectively.

For the first model, the dimension will be as follows:

Length/2 by width/2

94/2 by 50/2 = 47 feet by 25 feet.

For the second model, the dimension will be as follows:

Length/16 by width/16

The dimensions of the second model is 94/16 by 50/16 = 5.875 feet by 3.125 feet.

Since we are to solve for the area of the smallest scale model which is

5.875 feet by 3.125 feet.

Hence, area (A) = L× W

=5.875 × 3.125 feet.

= 18.359ft^2

4 0
3 years ago
The average depth of the ocean is about 3.8 km.width of 7600 km. calculate the ratio depth:width?
Gnoma [55]
A ratio is like a fraction, so since you are provided with the numbers its easy to put it into a ratio 
depth is 3.8km
width is 7600km
ratio is 3.8:7600
5 0
3 years ago
If x = a cosθ and y = b sinθ , find second derivative
Olin [163]

I'm guessing the second derivative is for <em>y</em> with respect to <em>x</em>, i.e.

\dfrac{\mathrm d^2y}{\mathrm dx^2}

Compute the first derivative. By the chain rule,

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}

We have

y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta

x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta

and so

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta

Now compute the second derivative. Notice that \frac{\mathrm dy}{\mathrm dx} is a function of \theta; so denote it by f(\theta). Then

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}

By the chain rule,

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}

We have

f=-\dfrac ba\cot\theta\implies\dfrac{\mathrm df}{\mathrm d\theta}=\dfrac ba\csc^2\theta

and so the second derivative is

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac ba\csc^2\theta}{-a\sin\theta}=-\dfrac b{a^2}\csc^3\theta

4 0
3 years ago
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