2 years ago

Will somebody help me with this ? Very much appreciated!! I’ll mark u as brainliest!

Mathematics

1 answer:

Mice21 [21]2 years ago

3 0

Answer:

Step-by-step explanation:

y = x - 1/4

x = 1/4 Put this value in for x

y = 1/4 - 1/4

y = 0

x = 13/4

y = 13/4 - 1/4 The denominators remain the same The numerators are subtracted.

y = (13 - 1)/4

y = 12/4

y = 3

If x is to = 2 then it must be changed so it's denominator is 4

2 = 8/4

y= 8/4 - 1/4

y = 7/4 or 1.75 or 1 3/4

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When y increases x increases

aniked [119]

Answer:

It happens only when they are inversely proportional to eachother.

X = 1/y

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3 years ago

Show that a sequence {sn} coverages to a limit L if and only if the sequence {sn-L} coverages to zero.

Andreyy89

Let ${s_n}_{n\in\Bbb N}$ be a sequence that converges to $L$ . This means for any $\varepsilon>0$ , there is some $N$ such that $|s_n-L|$ for all $n>N$ . From this inequality we see that $|(s_n-L)-0|$ , so it follows that $s_n-L\to0$ .

On the other hand, let ${s_n-L}$ be a sequence that converges to 0. This means $|(s_n-L)-0|$ for all large enough $n$ , and we get the simpler inequality for free, $|s_n-L|$ , so it follows that $s_n\to L$ .