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Gala2k [10]
1 year ago
9

Mr. McCracken and his friends are going on a golf trip. He wants to make golf bags to take on the trip. He has 24 golf trees and

36 golf balls. Each golf bag must have the same number of golf trees and the same number of golf balls. What are the possible numbers of golf bags Mr. McCracken can make so that the golf trees and golf balls are shared equally? Describe each possibility.
Mathematics
1 answer:
aleksandr82 [10.1K]1 year ago
5 0

There are in total Six cases possible - 1,2 ,3,4,6,12

Number of Golf Trees = 24

Number of Golf balls = 36

H.C.F [ Highest common factor] -

⇒ 24 = 2 * 2 * 2 * 3 * 1

⇒ 36 = 2 * 2 * 3 * 3 * 1

Case 1 ) Bag = 1 [ Total Golf trees = 24]

                           [ Golf trees = 36] [in each bag]

Case 2 ) Total Bags = 2

[Golf balls = 18]

[ Golf trees = 12] [in each bag]

Case 3 ) Total Bags = 3

Golf balls = 8

Golf trees = 10  [in each bag]

Case 4 ) Total Bags = 4

Golf balls = 9

Golf trees = 8 [in each bag]

Case 5 ) Total Bags = 6

Golf trees = 4

Golf balls = 6 [ in each bag ]

Case 6 ) Total Bags = 12

Golf balls = 3

Golf trees = 2 [ in each bag ]

Hence, There are in total Six cases possible - 1,2 ,3,4,6,12

To understand more about probability :

brainly.com/question/25870256

#SPJ9

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The area of the triangle formed by x− and y− intercepts of the parabola y=0.5(x−3)(x+k) is equal to 1.5 square units. Find all p
Juliette [100K]

Check the picture below.


based on the equation, if we set y = 0, we'd end up with 0 = 0.5(x-3)(x-k).

and that will give us two x-intercepts, at x = 3 and x = k.

since the triangle is made by the x-intercepts and y-intercepts, then the parabola most likely has another x-intercept on the negative side of the x-axis, as you see in the picture, so chances are "k" is a negative value.

now, notice the picture, those intercepts make a triangle with a base = 3 + k, and height = y, where "y" is on the negative side.

let's find the y-intercept by setting x = 0 now,


\bf y=0.5(x-3)(x+k)\implies y=\cfrac{1}{2}(x-3)(x+k)\implies \stackrel{\textit{setting x = 0}}{y=\cfrac{1}{2}(0-3)(0+k)} \\\\\\ y=\cfrac{1}{2}(-3)(k)\implies \boxed{y=-\cfrac{3k}{2}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of a triangle}}{A=\cfrac{1}{2}bh}~~ \begin{cases} b=3+k\\ h=y\\ \quad -\frac{3k}{2}\\ A=1.5\\ \qquad \frac{3}{2} \end{cases}\implies \cfrac{3}{2}=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)


\bf \cfrac{3}{2}=\cfrac{3+k}{2}\left( -\cfrac{3k}{2} \right)\implies \stackrel{\textit{multiplying by }\stackrel{LCD}{2}}{3=\cfrac{(3+k)(-3k)}{2}}\implies 6=-9k-3k^2 \\\\\\ 6=-3(3k+k^2)\implies \cfrac{6}{-3}=3k+k^2\implies -2=3k+k^2 \\\\\\ 0=k^2+3k+2\implies 0=(k+2)(k+1)\implies k= \begin{cases} -2\\ -1 \end{cases}


now, we can plug those values on A = (1/2)bh,


\bf \stackrel{\textit{using k = -2}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-2)\left(-\cfrac{3(-2)}{2} \right)\implies A=\cfrac{1}{2}(1)(3) \\\\\\ A=\cfrac{3}{2}\implies A=1.5 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{using k = -1}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-1)\left(-\cfrac{3(-1)}{2} \right) \\\\\\ A=\cfrac{1}{2}(2)\left( \cfrac{3}{2} \right)\implies A=\cfrac{3}{2}\implies A=1.5

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Number of pages = 9/1 = 9

2 - digit page number :

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Number of pages = 180/2 = 90

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588 - (9 + 180) = 588 - 189 = 399

Number of pages with 3 - digits = 399/3 = 133

Therefore, the total number of pages the book has is :

(9 + 90 + 133) = 232 pages

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