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wariber [46]
2 years ago
12

Which of the following Lacrosse players has a better reliability?

Mathematics
1 answer:
g100num [7]2 years ago
7 0
Gavin has better reliability
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3 years ago
1. What is the area of a circle with a diameter of 12.6 in.?
Rudik [331]

Answer:

Part 1) A=124.63\ in^{2}

Part 2) Option b

Part 3)  As n increases,  ns get closer to 2\pi r  

Part 4) Option c 16\pi\ ft^{2}

Part 5) Option b. 12.25\pi\ m^{2}

Step-by-step explanation:

Part 1) What is the area of a circle with a diameter of 12.6 in.?

we know that

the area of a circle is equal to

A=\pi r^{2}    

we have

r=12.6/2=6.3\ in -----> the radius is half the diameter

substitute the values

A=(3.14)(6.3^{2})=124.63\ in^{2}  

Part 2) Which explanation can be used to derive the formula for the circumference of a circle?

First find the relationship of the circumference to its diameter by finding that the length of the diameter wraps around the length of the circumference approximately ​ π ​ times.

Use this relationship to write an equation showing the ratio of circumference to diameter equaling ​ π

so

\frac{C}{D}=\pi

Rearrange the equation to solve for the circumference

C=\pi D

Substitute the diameter for 2 times the radius

D=2r

C=2\pi r                      

Part 3)  we know that

If n increases

then

the product ns get closer to the circumference of the circle

so

the circumference of a circle is equal to C=2\pi r  

therefore

As n increases,  ns get closer to 2\pi r        

Part 4) What is the area of a circle whose radius is 4 ft?

we know that

the area of a circle is equal to

A=\pi r^{2}

we have

r=4\ ft

substitute the values

A=(\pi)(4^{2})=16\pi\ ft^{2}

Part 5) The circumference of a circle is 7π m.

What is the area of the circle?

we know that

The circumference of a circle is equal to

C=2\pi r

we have

C=7\pi\ m

substitute and solve for r

7\pi=2\pi r

r=3.5\ m

Find the area of the circle

the area of a circle is equal to

A=\pi r^{2}

substitute

the area of a circle is equal to

A=\pi (3.5^{2})=12.25\pi\ m^{2}

3 0
3 years ago
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