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Alex787 [66]
3 years ago
6

The area of a rectangle room is 750 square feet. The width of the room is 5 feet less than the length of the room. Which equatio

n can be used to solve for Y the length of the room
Mathematics
1 answer:
olga_2 [115]3 years ago
5 0
By definition, the area of a rectangle is given by:
 A = (width) * (long)
 Substituting values we have:
 750 = (y-5) * (y)
 Rewriting we have:
 750 = y ^ 2 - 5y
 y ^ 2 - 5y - 750 = 0
 Answer:
 
An equation that can be used to solve for And the length of the room is:
 
y ^ 2 - 5y - 750 = 0
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<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

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2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2

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  • Let us assume that the result is true for n=k

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<u>To prove:</u>  2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Let us take n=k+1

Hence, we have:

2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)

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Hence, we have:

2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2

Also, we know that:

-2

(

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2^{(k+1)+1}-2^{(k+1)-1}>0  )

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2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

2^n\leq 2^{n+1}-2^{n-1}-1  where n is a positive integer.

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