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Flura [38]
3 years ago
5

Y = x-1 Given x = 5 y =

Mathematics
2 answers:
Flura [38]3 years ago
4 0

Answer:

When x=5, y=4

Step-by-step explanation:

Whenever you have to solve these kinds of equations, all you have to do is substitute what you already have into it.

<u>The equation you've been given is:</u>

y = x - 1

They want you to find what y is, when x = 5

<u>So substitute in x=5:</u>

y = 5 - 1

y=4

Hope that helped : )

Firdavs [7]3 years ago
3 0
The answer would be 4 bc 5 minus 1 is 4. x = 4. y = 4.
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Assume L=1.5W, where W=width, L= Length of the triangle.

Therefore,

Perimeter (P) = 2(W+L) = 2(W+1.5W) = 2(2.5W) = 5W=20 => W=20/5 = 4 in

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Step-by-step explanation:

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3 years ago
Mathematical induction, prove the following two statements are true
adelina 88 [10]
Prove:
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____________________________________________

Base Step: For n=1:
n\left(\frac12\right)^{n-1}=1\left(\frac12\right)^{0}=1
and
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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}}
assumed to be true.

--------------------------------------------------------------------------

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,
=4-\dfrac{k+2}{2^{k-1}}+\dfrac{(k+1)}{2^{k}}

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,
=4-\dfrac{k+3}{2^{k}}

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,
=4-\dfrac{(k+1)+2}{2^{(k-1)+1}}
3 0
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