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Lynna [10]
3 years ago
13

Jing jing works in a store and she spends some of her time organizing the stationary section. If there are nine different items

on the rack, how many ways can she organize the rack so that the two most expensive items are not together?
Mathematics
1 answer:
Alexus [3.1K]3 years ago
4 0

Answer:

282240 ways

Step-by-step explanation:

number of items = 9 different items

Ranking based on worth

<u>determine number of ways to arrange the 9 items such that the 2 most expensive are apart</u>

first step: consider the 2expensive items as 1

∴ number of permutation = [ ( 9 - 2 ) + 1 ]!  * 2!

                                          = [ 8 ] ! * 2!  = 80640

Total number of permutation/arrangement = 9! = 362880

hence to arrange the 9 items without having the two most expensive items together

= 362,880 - 80640

= 282240 ways

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Yes; Opposite sides are congruent, and diagonals are congruent.

Step-by-step explanation:

we have

A(4, -7), B(4, -2), C(0, -2), D(0, -7)

we know that

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d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

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<u><em>Find the distance BC</em></u>

substitute the values

d=\sqrt{(-2+2)^{2}+(0-4)^{2}}

d=\sqrt{(0)^{2}+(-4)^{2}}

BC=4\ units

<u><em>Find the distance CD</em></u>

substitute the values

d=\sqrt{(-7+2)^{2}+(0-0)^{2}}

d=\sqrt{(-5)^{2}+(0)^{2}}

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<u><em>Find the distance AD</em></u>

substitute the values

d=\sqrt{(-7+7)^{2}+(0-4)^{2}}

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Find the length of the diagonals

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substitute the values

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d=\sqrt{(5)^{2}+(-4)^{2}}

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substitute the values

d=\sqrt{(-7+2)^{2}+(0-4)^{2}}

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Compare the length of the diagonals

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therefore

Diagonals are congruent

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