Paul is making a ball-toss game for his club booth at the fair. He wants to make the circumference of the holes at least 3 inches greater but not more than 4 inches greater than the circumference of the ball. One person suggests that Paul can make the diameter of the hole 1 inch greater than that of the ball. Another suggests the diameter should be 2inches greater. Which suggestion should Paul choose? Explain

Step-by-step explanation:

we are going to assume a diameter of the hole has 6in

the radius of the ball is 3in

we looking at a range of the hole greater by (3<4)

we know that the circumference of a circle is given as

the circumference of the ball is

C=2πr

C=2*3.142*3

C=18.85in

The circumference of the hole is ( 1 inch greater )

C=2πr

say the diameter is 6in plus 1in = 7in

the radius =3.5in

C=2*3.142*3.5

C=21.99

compare with the ball the difference is 21.99-18.85

=3.14 which is in the range

The circumference of the hole is ( 2inches greater. )

C=2πr

say the diameter is 6in plus 2in = 8in

the radius =4in

C=2*3.142*4

C=25.13in

compare with the ball the difference is 25.13-18.85

=6.28 which is out of  the range

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hope this .. : d

something that characterises a right triangle is the fact that we may use pythagoras' theorem to find any one of its sides given that we have the length of the other two sides. here, we have the length of the hypotenuse, √(87) yd, and one other side, √(23) yd, thus we can easily find the length of the remaining side using pythagoras' theorem as defined below:

c^2 = a^2 + b^2, where c is the length of the hypotenuse, and a and b are the lengths of the other two sides.

thus, given that c = √(87) and a = √(23), we may substitute these into the formula we defined above to get:

c^2 = a^2 + b^2

(√87)^2 = (√23)^2 + b^2 (substitute c = √87 and a = √23 into the equation)

87 = 23 + b^2 (evaluate (√87)^2 = 87 and (√23)^2 = 23 - remember that the square root of a number, squared will equal that very same number)

64 = b^2 (subtract 23 from both sides)

√64 = b (take the square root of both sides)

8 = b (evaluate √64 = 8)

therefor, the length of the unknown leg in the right triangle is 8 yd (first answer).