Asked by Alexandria
                the distance between any two bases on a baseball diamond is 90 feet.What is the best approximation for the direct distance from home plate to second base?
A baseball diamond's four bases form a square with 90 foot sides. A right triangle is formed by home plate, first and second base. The distance from second base to home is the hypotenuse of that triangle.
Apply the Pythagorean theorem
a^2 + b^2 = c^2 to the right triangle and you will have the answer. In this case, a = b = 90. Solve for c.
            
        A baseball diamond's four bases form a square with 90 foot sides. A right triangle is formed by home plate, first and second base. The distance from second base to home is the hypotenuse of that triangle.
Apply the Pythagorean theorem
a^2 + b^2 = c^2 to the right triangle and you will have the answer. In this case, a = b = 90. Solve for c.
Answers
                    Answered by
            michael
            
    this is not the question i was looking for!!! >.<
  
i wanted a related rates question!!!!!!
i have a test on it!!!!!
omg ><
    
i wanted a related rates question!!!!!!
i have a test on it!!!!!
omg ><
                    Answered by
            Anonymous
            
    127.28
    
                    Answered by
            Kim
            
    Which statement is true about the following pair of triangles?
Square with one side 30 am done side 60 and a smaller square one side 6 and one side 3
    
Square with one side 30 am done side 60 and a smaller square one side 6 and one side 3
                    Answered by
            goo-goo-gaa-gaa
            
    93 ft
    
                    Answered by
            cat loaf
            
    this was literally posted before i was born 
    
                    Answered by
            borgor
            
    d^2 = a^2 + b^2. 
a=90, b=90
d^2 = 90^2 + 90^2 >>> d^2 = 2 * 90^2
d = 90 sq.rt.2 ft.
    
a=90, b=90
d^2 = 90^2 + 90^2 >>> d^2 = 2 * 90^2
d = 90 sq.rt.2 ft.
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