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Butoxors [25]
4 years ago
10

Amrita thinks of a number, she doubles it, adds 9, divides by three and then subtracts one.

Mathematics
2 answers:
FinnZ [79.3K]4 years ago
5 0

Answer:

x = 6

Step-by-step explanation:

<em>Let the number be x</em>

<u><em>Now let's make an equation for it:</em></u>

=> \frac{2x+9}{3} -1 = x

=> \frac{2x+9}{3} = x+1

<em>Multiplying both sides by 3</em>

=> 2x+9 = 3(x+1)

=> 2x+9 = 3x+3

=> 3x-2x = 9-3

=> x = 6

wariber [46]4 years ago
3 0

Answer:

I believe your answer would be 6.

Step-by-step explanation:

I have set up an equation that represents the situation.

Letting the unknown number being 'n':

\frac{2n+9}{3} -1=n

\frac{2n+9}{3} -1=n\\\\\frac{2n+9}{3}-1+1=n+1\\\\\frac{2n+9}{3}=n+1\\\\\frac{2n+9}{3}*3=3(n+1)\\\\2n+9=3n+3\\\\2n+9-9=3n+3-9\\\\2n=3n-6\\\\2n-3n=3n-3n-6\\\\-n=-6\\\\\frac{-n}{-1}=\frac{-6}{-1}\\\\  \boxed{n=6}

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Induction Hypothesis: Assume true for n=k. Meaning:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}=4-\dfrac{k+2}{2^{k-1}}
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Induction Step: For n=k+1:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}

by our Induction Hypothesis, we can replace every term in this summation (except the last term) with the right hand side of our assumption.
=4-\dfrac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^{k}

From here, think about what you are trying to end up with.
For n=k+1, we WANT the formula to look like this:
1+2\left(\frac12\right)+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}=4-\dfrac{(k+1)+2}{2^{(k+1)-1}}

That thing on the right hand side is what we're trying to end up with. So we need to do some clever Algebra.

Combine the (k+1) and 1/2, put the 2 in the bottom,
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We want to end up with a 2^k as our final denominator, so our middle term is missing a power of 2. Let's multiply top and bottom by 2,
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Distribute the -2 and combine the fractions together,
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Combine like-terms,
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pull the negative back out,
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And ta-da! We've done it!
We can break apart the +3 into +1 and +2,
and the +0 in the bottom can be written as -1 and +1,
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3 0
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