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NeTakaya
3 years ago
14

middle" class="latex-formula"> ÷ 39


show the work with the answer!
Mathematics
1 answer:
vfiekz [6]3 years ago
4 0

Answer:

\frac{1}{30}

Step-by-step explanation:

Change the mixed number into an improper fraction

1 \frac{3}{10} = \frac{13}{10}, hence

\frac{13}{10} ÷ 39

To divide, leave the first fraction, change division to multiplication and turn 39 upside down, that is

= \frac{13}{10} × \frac{1}{39}

cancel the 13 and 39 by dividing both by 13

= \frac{1}{10(3)} = \frac{1}{30}


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Can you define f(0, 0) = c for some c that extends f(x, y) to be continuous at (0, 0)? If so, for what value of c? If not, expla
Ahat [919]

(i) Yes. Simplify f(x,y).

\displaystyle \frac{x^2 - x^2y^2 + y^2}{x^2 + y^2} = 1 - \frac{x^2y^2}{x^2 + y^2}

Now compute the limit by converting to polar coordinates.

\displaystyle \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} = \lim_{r\to0} \frac{r^4 \cos^2(\theta) \sin^2(\theta)}{r^2} = 0

This tells us

\displaystyle \lim_{(x,y)\to(0,0)} f(x,y) = 1

so we can define f(0,0)=1 to make the function continuous at the origin.

Alternatively, we have

\dfrac{x^2y^2}{x^2+y^2} \le \dfrac{x^4 + 2x^2y^2 + y^4}{x^2 + y^2} = \dfrac{(x^2+y^2)^2}{x^2+y^2} = x^2 + y^2

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\dfrac{x^2y^2}{x^2+y^2} \ge 0 \ge -x^2 - y^2

Now,

\displaystyle \lim_{(x,y)\to(0,0)} -(x^2+y^2) = 0

\displaystyle \lim_{(x,y)\to(0,0)} (x^2+y^2) = 0

so by the squeeze theorem,

\displaystyle 0 \le \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} \le 0 \implies \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} = 0

and f(x,y) approaches 1 as we approach the origin.

(ii) No. Expand the fraction.

\displaystyle \frac{x^2 + y^3}{xy} = \frac xy + \frac{y^2}x

f(0,y) and f(x,0) are undefined, so there is no way to make f(x,y) continuous at (0, 0).

(iii) No. Similarly,

\dfrac{x^2 + y}y = \dfrac{x^2}y + 1

is undefined when y=0.

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2 years ago
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