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Anna71 [15]
3 years ago
5

Carl has one kind of bread

Mathematics
2 answers:
NNADVOKAT [17]3 years ago
6 0
What else is in the question this isn't easy to answer
love history [14]3 years ago
5 0
And??? Does Carl eat his one kind of bread??? What's the question?
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Answer: location

Step-by-step explanation:

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What is 5/8 + 3/4 using these sticks
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3/4 =6/8 so you have 3/4slices of pizza cut the 3 of them in half

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Three friends each have some ribbon. Carol has 54 inches of ribbon, Tino has 4.5 feet of ribbon, and Baxter has 6.5 yards of rib
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Mathematical induction, prove the following two statements are true
adelina 88 [10]
Prove:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+n\left(\frac12\right)^{n-1}=4-\dfrac{n+2}{2^{n-1}}
____________________________________________

Base Step: For n=1:
n\left(\frac12\right)^{n-1}=1\left(\frac12\right)^{0}=1
and
4-\dfrac{n+2}{2^{n-1}}=4-3=1
--------------------------------------------------------------------------

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

Distribute the -2 and combine the fractions together,
=4+\dfrac{-2k-4+(k+1)}{2^{k}}

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

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
3 years ago
Please help.. (This is meant to be infinite, so I can't use a calculator)
Alex17521 [72]

Answer:

5

Step-by-step explanation:

→ Let x = √20 + √20 .....

x

→ Square both sides

x² = 20 + √20 +√20 + √20

→ Replace √20 +√20 + √20 with x

x² = 20 + x

→ Move everything to the left hand side

x² - 20 - x = 0

→ Factorise

( x + 4 ) ( x - 5 )

→ Solve

x = -4 , 5

→ Discard negative result

x = 5

7 0
2 years ago
Read 2 more answers
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