Suppose a certain baseball diamond is a square 80 feet on a side. The pitching rubber is located 54.5 feet from home plate on a line joining home plate and second base. ​(a) How far is it from the pitching rubber to first​ base?

​(b) How far is it from the pitching rubber to second​ base?

​(c) If a pitcher faces home​ plate, through what angle does he need to turn to face first​ base?

answer: just subtact the units and then add the non negative : happy to

step-by-step explanation:

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x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15