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kupik [55]
2 years ago
12

BRALIEST IF CORRECT :))

Mathematics
2 answers:
statuscvo [17]2 years ago
8 0

Answer:

-9

Step-by-step explanation:

because the - is invisible and is square root

yuradex [85]2 years ago
4 0
A. Is the right answer I think I have e does this one before back in5th so I’m sure it’s A.
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Ben fills 42 pints of juice in 12 bottles. Six bottles have a capacity of 3 pints each. The remaining 6 bottles are all of the s
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Answer:

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5 0
3 years ago
1. Find the derivative with respect to x of x +1/x from first principle. <br>​
Murrr4er [49]

If you mean f(x)=x+\frac1x, then the derivative is

\displaystyle f'(x) = \lim_{h\to0} \frac{\left(x+h+\frac1{x+h}\right) - \left(x+\frac1x\right)}h \\\\ = \lim_{h\to0} \frac{(x+h)-x}h + \lim_{h\to0} \frac{\frac1{x+h} - \frac1x}h \\\\ = \lim_{h\to0} \frac hh + \lim_{h\to0} \frac{x-(x+h)}{hx(x+h)} \\\\ = \lim_{h\to0} 1 - \lim_{h\to0} \frac h{hx(x+h)} \\\\ = 1 - \lim_{h\to0} \frac1{x(x+h)} \\\\ = \boxed{1 - \frac1{x^2}}

If you mean f(x) = \frac{x+1}x = 1 + \frac1x, we know from above that

\displaystyle \left(\frac1x\right)' = \lim_{h\to0} \frac{\frac1{x+h}-\frac1x}h = -\frac1{x^2}

which leaves the constant term, whose derivative is

\displaystyle (1)' = \lim_{h\to0}\frac{1 - 1}h = 0

and so

f'(x) = -\dfrac1{x^2}

6 0
2 years ago
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