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Yuri [45]
3 years ago
9

The length of a rectangular garden is three feet less than twice it’s width. If the perimeter of the garden is 42 feet, what is

the length?
Mathematics
1 answer:
zzz [600]3 years ago
6 0
<h3><u>The length is equal to 13.</u></h3><h3><u>The width is equal to 8.</u></h3>

l = 2w - 3

2l + 2w = 42

Because we have a value for l, we can plug it into the second equation to solve for w.

2(2w - 3) + 2w = 42

Distributive property.

4w - 6 + 2w = 42

Combine like terms.

6w - 6 = 42

Add 6 to both sides.

6w = 48

Divide both sides by 6.

w = 8

Now that we have a value for w, we can solve for the exact value of l.

l = 2(8) - 3

l = 16 - 3

l = 13



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I run a book club with n people, not including myself. Every day, for 365 days, I invite three members in the club to review a b
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<h3>Answer:   15</h3>

========================================================

Explanation:

The order doesn't matter. A group like {A,B,C} is the same as {B,A,C}.

All that matters is the overall group rather than the positioning of the members.

We'll use the nCr combination formula since order doesn't matter.

The value of n is unknown, but we know that r = 3 members are to be selected.

Let's pick a value for n at random. Let's say n = 10.

Plug n = 10 and r = 3 into the nCr formula below.

n C r = \frac{n!}{r!(n-r)!}\\\\10 C 3 = \frac{10!}{3!*(10-3)!}\\\\10 C 3 = \frac{10!}{3!*7!}\\\\10 C 3 = \frac{10*9*8*7!}{3!*7!}\\\\ 10 C 3 = \frac{10*9*8}{3!}\\\\ 10 C 3 = \frac{10*9*8}{3*2*1}\\\\ 10 C 3 = \frac{720}{6}\\\\ 10 C 3 = 120\\\\

Unfortunately we don't reach 365 or larger.

Let's try n = 11

n C r = \frac{n!}{r!(n-r)!}\\\\11 C 3 = \frac{11!}{3!*(11-3)!}\\\\11 C 3 = \frac{11!}{3!*8!}\\\\11 C 3 = \frac{11*10*9*8!}{3!*8!}\\\\ 11 C 3 = \frac{11*10*9}{3!}\\\\ 11 C 3 = \frac{11*10*9}{3*2*1}\\\\ 11 C 3 = \frac{990}{6}\\\\ 11 C 3 = 165\\\\

We're still under our target. The good news is that the nCr value is increasing.

So the idea is to do trial and error with various values of n. Keep incrementing n until nCr = nC3 is equal to 365 or larger.

Here's a table of values where r = 3 the entire time

\begin{array}{|c|c|} \cline{1-2}\text{n} & \text{nCr}\\\cline{1-2}10 & 120\\\cline{1-2}11 & 165\\\cline{1-2}12 & 220\\\cline{1-2}13 & 286\\\cline{1-2}14 & 364\\\cline{1-2}15 & 455\\\cline{1-2}\end{array}

The nCr values are also found in Pascal's Triangle. Each of those values are the fourth entry of each row.

When n = 14, we have nCr = 364 which is very close. We're one short unfortunately.

So we have to go for <u>n = 15</u> instead. This makes the nCr value well over 365 of course, but it guarantees that you'll have plenty of trios to choose from such that no group of three is repeated. Unfortunately some trios will be left out.

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Step-by-step explanation:

divide 2/3 in half because you're going from 2 dozen to 1 dozen

2/3 ÷ 2 equals 2/3 x 1/2 which equals 1/3

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