A.
Answer:
~44.65 ft
Explanation:
a = 25
b = 37
c = ?
c^2 = a^2 + b^2
So,
c = √a^2+b^2
c = √(25)^2+(37)^2
c = 44.6542271235 ft
the player is 44.65 ft from the hoop
B.
50 ft
a = 40
b = 30
c = √(40)^2+(30^2
c = √1600+900
c = √2500
c = 50 ft
1. about 7.87 feet away
2. About 8.37 yds
1. a^2+b^2=c^2
so 25+37= C^2
= 62
Square root of 62 = 7.87400787401
(About 7.87 ft away)
2. 40+30 = C^2
Square root of 70 = 8.36660026534
About 8.37
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