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Alborosie
2 years ago
7

Pls answer correctly I’ll give you brainliest

Mathematics
1 answer:
baherus [9]2 years ago
7 0

Answer:

y=-1x+-3

Step-by-step explanation:

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Given the figure below, find the values of x and z
Ratling [72]
(13x-85)^o=97^o\\\\13x-85=97\ \ \ |+85\\\\13x=182\ \ \ |:13\\\\x=14

z^o+97^o=180^o\ \ \ |-97^o\\\\z^o=83^o

Answer: x = 14; z = 83
4 0
3 years ago
Find the number to which the sequence {(3n+1)/(2n-1)} converges and prove that your answer is correct using the epsilon-N defini
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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
HELP HELP WHATS THE ANSWER TO THIS HELP YOU'LL GET 50+ BRAINLIEST POINTS AND BRAINLIEST HELP PLS
saveliy_v [14]

Answer:

It says not a conic section

Step-by-step explanation:

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0.024

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3 years ago
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