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Aleks [24]
3 years ago
14

Solve for x. I need help :,) thank you if you help!

Mathematics
1 answer:
natta225 [31]3 years ago
5 0

Answer: if its a multiple of a 345 triangle then its congruent x is 36 if othersides ate multiples of 20 ie

Sides are

60 80 100 is a 345 right triangle

Step-by-step explanation:

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Answer:

The answer is within the range of 0.1 - 0.9.

Step-by-step explanation:

For example, 0.5 could be the answer.

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Double-Angle and Half-Angle Identiies [See Attachment] Question 4
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Please make it brainleist

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20,18,13,17,18 find the mean,median and mode​
Degger [83]
<h3>Answer:</h3>

Mean: 17.2

Median: 18

Mode: 18

<h3>Solution:</h3>

Mean: The sum of all observation divided by several observation

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Find the number to which the sequence {(3n+1)/(2n-1)} converges and prove that your answer is correct using the epsilon-N defini
Nat2105 [25]
By inspection, it's clear that the sequence must converge to \dfrac32 because

\dfrac{3n+1}{2n-1}=\dfrac{3+\frac1n}{2-\frac1n}\approx\dfrac32

when n is arbitrarily large.

Now, for the limit as n\to\infty to be equal to \dfrac32 is to say that for any \varepsilon>0, there exists some N such that whenever n>N, it follows that

\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|

From this inequality, we get

\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|=\left|\dfrac{(6n+2)-(6n-3)}{2(2n-1)}\right|=\dfrac52\dfrac1{|2n-1|}
\implies|2n-1|>\dfrac5{2\varepsilon}
\implies2n-1\dfrac5{2\varepsilon}
\implies n\dfrac12+\dfrac5{4\varepsilon}

As we're considering n\to\infty, we can omit the first inequality.

We can then see that choosing N=\left\lceil\dfrac12+\dfrac5{4\varepsilon}\right\rceil will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that N\in\mathbb N.
6 0
3 years ago
How do do the work for 12c+6c=36
BlackZzzverrR [31]
Apologies for the bad handwriting.

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3 years ago
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