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.
127.28
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
93 ft
this was literally posted before i was born
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.
To find the best approximation for the direct distance from home plate to second base on a baseball diamond, we can use the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.
In this case, we have a right triangle with the lengths of both legs equal to 90 feet, which are the sides of the square formed by the bases. Let's label one leg as "a" and the other leg as "b". According to the Pythagorean theorem, we need to solve for the length of the hypotenuse, which we'll call "c".
Using the Pythagorean theorem formula: a^2 + b^2 = c^2, we can substitute the lengths of the legs:
90^2 + 90^2 = c^2
Simplifying this equation, we have:
8100 + 8100 = c^2
16200 = c^2
Now, we can find the square root of both sides to solve for "c":
c = √16200
Using a calculator, we get:
c ≈ 127.28
Therefore, the best approximation for the direct distance from home plate to second base is approximately 127.28 feet.
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i wanted a related rates question!!!!!!
i have a test on it!!!!!
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