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sergiy2304 [10]
3 years ago
8

Coach Stevens needs to purchase sprinklers to water the baseball field. The standard distance between bases is 90 feet and the i

nfield is a perfect square.
Coach Stevens found one sprinkler that sprays a maximum distance of 50 feet.


If Coach Stevens placed the sprinkler on the pitcher's mound, would the water spray so that the entire infield was watered? Justify your response.

Mathematics
1 answer:
egoroff_w [7]3 years ago
6 0
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.


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Find X if m/_ UVH=50°, m/_HVW=25x-5, and m/_ UVW = 35x-5.​
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Step-by-step explanation:

first you have to solve for x

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Why is ac smaller than ad if they both look very close
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Let's begin by listing out the information given to us:

We start out by observing that Triangles MKR & ACD are similar or proportional

\begin{gathered} MK=21;AC=\text{?} \\ MR=24;AD=28\frac{4}{5} \\ KR=CD=\text{?} \end{gathered}

We will solve for the missing side by using the similar triangle theorem. This is shown below:~

\begin{gathered} \Delta MKR\approx\Delta ACD \\ \frac{MK}{AC}=\frac{MR}{AD} \\ \frac{21}{AC}=\frac{24}{28\frac{4}{5}} \\ \text{Cross multiply, we have:} \\ 24\cdot AC=28\frac{4}{5}\cdot21 \\ AC=\frac{28\frac{4}{5}\cdot21}{24}=25\frac{1}{5} \\ AC=25\frac{1}{5} \end{gathered}

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