105 quarters are lying on a flat surface with their edges in contact. They are just contained by a frame in?

Question

the form of an equilateral triangle. If the inside perimeter of the frame is 42 inches, find the radius of a quarter.

2) A rectangular sheet of paper ABCD is three inches in width and four inches in length. The paper is folded so that the two diagonally opposite corners A and C coincide. Determine the length of the crease in the paper.

3) If x,y, and z are unequal real numbers whose sum is zero and whose product is two, determine the value of x^3+y^3+z^3.

4)Among grandfather's papers an old bill was found:"72 Turkeys $)_67.9_." The first and last digits of the number that represented the totaly price of the turkeys are replaced here with blanks as they had faded and are now illegible. What are the msising digits?

Thanks so much!

Answer 1

Hi,

Here are 3 of 4:

1) Correction:
In order to form an equilateral triangle each row of quarters has one more quarter than on the previous row. In order for the total to be 105 quarters, the longest row will have 14 quarters. (This is true because the arithmetic sequence from 1 to 14 has a sum of S = n/2(a + l) = 14/2(1 + 14) = 7(15) = 105.) Since the perimeter is 42 inches, then each side of the equilateral triangle is 42 inches. Each side has 14 quarters in a row, but there has to be the additional room to make the corners of the triangle. Let r = the radius of a quarter. Then in the row of 14 quarters, there are 13 spaces of "2r" from center to center of the end quarters. In each corner of the triangle there is a 30-60-90° triangle formed where the 30° side is the radius of the quarter. The distance from directly below its center on the edge of the equilateral triangle to the corner of the equilateral triangle is the 60° side, which is r√3
inches long. Since this is on both ends, the total side length of 14 inches equals 2r√3 + 13(2r). Solving this for "r" gives:
..............._
14 = 2r√3 + 13(2r)
..............._
14 = 2r√3 + 26r
.........._
7 = r√3 + 13r
..........._
7 = r(√3 + 13)

...7
---------- = r
√3 + 13

Accounting for the space in the corners, this is the radius of the quarter.




2) From A to D is the 4 inch length and from D to C is the 3 inch width. If corner A is folded up onto point C, then part of the 4 inch side is folded up diagonally to point C and the rest of that side doesn't move. If the piece of AD remaining horizontal has a length of x inches. then the diagonal piece of AD that was folded up has a length of 4 - x inches. These form a right triangle with legs of x and 3 inches and a hypotenuse of 4 - x inches. So x² + 3² = (4 - x)². This multiplies out to x² + 9 = 16 - 8x + x². So, 9 = 16 - 8x and 8x = 7, so x = 7/8. That means the hypotenuse, which is the fold in the paper has a length of 4 - x or 4 - 7/8 to be 3 1/8" long. (3.125") These are the sides of a 7-24-25 right triangle as 7/8, 24/8 or 3, and 25/8 or 3 1/8.

3) Tony's answer for this is correct and well done.

4) Each turkey cost $5.11 because if you look at the equation y = 72x where x increases by.01, the only answer you get with the correct digits y67.9z is 72 x 5.11 or $367.92.

I hope that helps!! :-)

Answer 2

q1
105=1+2+3+4+5+6+7+8+9+10+11+12+13+14
=15*7
each side has 14 coins.
42/3=14=14*d
d=1, r=0.5

q2
diagonal of paper=sqrt(4^2+3^2)
=sqrt(25)
=5
when folded from 1 end of a diagonal to the other, we get an isoceles triangle with base of crease and height of half the diagonal. the angle btw base and edge is equal to the 1 of the btw of diagonal and width.
crease=2*3*5/4*2=3.75

Answer 3

3) We have x + y + z = 0 and xyz = 2. Because their sum is 0, we know that they cannot all have the same sign, and since their product is positive, we know two are negative and one is positive. Let's say x < 0, y < 0, anf z > 0. From the first equation we have z = -x - y, and from the second we have
xy(-x - y) = 2, or -x^2y - xy^2 = 2.

Now x^3 + y^3 + z^3 = x^3 + y^3 + (-x - y)^3 = -3x^2y -3xy^2 =
3(-x^2y - xy^2) = 3*2 = 6.

Answer 4

Last one: 72 turkeys at $5.11 a piece = $367.92