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Mathematics

1 answer:

leonid [27]2 years ago

5 0

I think the answer is D.

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A regulation baseball field measures 90 feet between each base. What is the distance around the bases in yards?

Naddika [18.5K]

Home to 1st base = 90 ft
1st base to 2nd base = 90 ft
2nd base to 3rd base = 90 ft
3rd base to home = 90 ft
so we have a total of : (90 * 4) = 360 ft

there are 3 ft in 1 yd....
so in 360 ft, there are : 360/3 = 120 yds <==

or u could just figure it between each base....90 ft = (90/3) = 30 yds per base...and u have 4 bases....so that (4 * 30) = 120 yds

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7+3√5/3+√5 - 7-3√5/3-√5 = a + b√5

daser333 [38]

<h3>Given:-</h3>

$\\ \sf \implies\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } = a + \sqrt{5} b \\$

The value of a and b

<h3>Solution:-</h3>

$\\ \sf \implies\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } = a + \sqrt{5} b \\$

$\\ \sf \implies\frac{( \: 7 + 3 \sqrt{5} \: \: ( 3 - \sqrt{5}) \: \: - 7 - 3 \sqrt{5} \: \: ( 3 + \sqrt{5}) \: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{( \: 21 - 7 \sqrt{5} \: + 9 \sqrt{5} - 15) \: \: - ( \: 21 + 7 \sqrt{5} \: - 9 \sqrt{5} + 15)\: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$