If you need the geometry and trigonometry I'll post those. Instead I found an answer at this link: http://jwilson.coe.uga.edu/EMAT6450/Class%20Projects/Scarpelli/Scarpelli_MathematicsBaseballActivity.....
The distance from home plate to the pitcher's mound is 60.5 feet and from home plate to second base is <span>127.28 feet.
Pitcher's Mound to 2nd base = </span><span>
<span>
<span>
66.78</span> </span>f</span>eet
Given those distances, we KNOW a 50 foot sprinkler will NOT reach home plate and second base from the pitcher's mound.
I didn't figure out the pitcher's mound to 1st or to 3rd, since the question is already answered.
Answer:
![\text{Exact area of the sidewalk}=40 \pi\text{ m}^2](https://tex.z-dn.net/?f=%5Ctext%7BExact%20area%20of%20the%20sidewalk%7D%3D40%20%5Cpi%5Ctext%7B%20m%7D%5E2)
![\text{Approximate area of the sidewalk}=125.6\text{ m}^2](https://tex.z-dn.net/?f=%5Ctext%7BApproximate%20area%20of%20the%20sidewalk%7D%3D125.6%5Ctext%7B%20m%7D%5E2)
Step-by-step explanation:
We have been given that at a zoo, the lion pen has a ring-shaped sidewalk around it. The outer edge of the sidewalk is a circle with a radius of 11 m. The inner edge of the sidewalk is a circle with a radius of 9 m.
To find the area of the side walk we will subtract the area of inner edge of the side walk of lion pen from the area of the outer edge of the lion pen.
, where r represents radius of the circle.
![\text{Exact area of the sidewalk}=\pi*\text{(11 m)}^2-\pi*\text{(9 m)}^2](https://tex.z-dn.net/?f=%5Ctext%7BExact%20area%20of%20the%20sidewalk%7D%3D%5Cpi%2A%5Ctext%7B%2811%20m%29%7D%5E2-%5Cpi%2A%5Ctext%7B%289%20m%29%7D%5E2)
![\text{Exact area of the sidewalk}=\pi*\text{121 m}^2-\pi*\text{81 m}^2](https://tex.z-dn.net/?f=%5Ctext%7BExact%20area%20of%20the%20sidewalk%7D%3D%5Cpi%2A%5Ctext%7B121%20m%7D%5E2-%5Cpi%2A%5Ctext%7B81%20m%7D%5E2)
![\text{Exact area of the sidewalk}=40 \pi\text{ m}^2](https://tex.z-dn.net/?f=%5Ctext%7BExact%20area%20of%20the%20sidewalk%7D%3D40%20%5Cpi%5Ctext%7B%20m%7D%5E2)
Therefore, the exact area of the side walk is ![40 \pi\text{ m}^2](https://tex.z-dn.net/?f=40%20%5Cpi%5Ctext%7B%20m%7D%5E2)
To find the approximate area of side walk let us substitute pi equals 3.14.
![\text{Approximate area of the sidewalk}=40*3.14\text{ m}^2](https://tex.z-dn.net/?f=%5Ctext%7BApproximate%20area%20of%20the%20sidewalk%7D%3D40%2A3.14%5Ctext%7B%20m%7D%5E2)
![\text{Approximate area of the sidewalk}=125.6\text{ m}^2](https://tex.z-dn.net/?f=%5Ctext%7BApproximate%20area%20of%20the%20sidewalk%7D%3D125.6%5Ctext%7B%20m%7D%5E2)
Therefore, the approximate area of the side walk is
.
Answer:
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Step-by-step explanation:
c I'm x go bc to ggzh it ti
Cage.....dog travel kit..... shelter.....pound.... it really depends on what you mean by dog kennel.
An equation which can be used to solve the given system of equations is 3x⁵ - 5x³ + 2x² - 10x + 4 = 4x⁴ + 6x³ - 11.
<u>Given the following data:</u>
y = 3x⁵ - 5x³ + 2x² - 10x + 4
y = 4x⁴ + 6x³ - 11
<h3>What is a system of equations?</h3>
A system of equations can be defined an algebraic equation that only has two (2) variables and can be solved simultaneoulsy.
Equating the given equations, we have:
y = y
3x⁵ - 5x³ + 2x² - 10x + 4 = 4x⁴ + 6x³ - 11
3x⁵ - 5x³ + 2x² - 10x + 4 - (4x⁴ + 6x³ - 11) = 0
3x⁵ - 5x³ + 2x² - 10x + 4 - 4x⁴ - 6x³ + 11 = 0
3x⁵ - 5x³ + 2x² - 10x + 4 - 4x⁴ - 6x³ + 11 = 0
3x⁵ - 4x⁴ - 11x³ + 2x² - 10x + 15 = 0
Read more on equations here: brainly.com/question/13170908