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VikaD [51]
3 years ago
9

Help with 6 and 7 I rlly need help

Mathematics
1 answer:
n200080 [17]3 years ago
7 0

Answer:

6) 295.2 feet squared. 7) 2, 4, 7, 8

Step-by-step explanation:

6)

a = 16.4 \times 18

a = 295.2

7)

728 \div 2 = 364

728 \div 4 = 182

728 \div 7 = 104

728  \div 8 = 91

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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
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(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

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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

and

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

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

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\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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