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frutty [35]
3 years ago
6

This is not a question, this is for who-ever needs it.

Mathematics
1 answer:
Eddi Din [679]3 years ago
5 0

Answer:

thank you so much!

Step-by-step explanation:

You might be interested in
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

and

\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.

5 0
2 years ago
Does anyone know these?? please help :(
omeli [17]
1- ABD
2- BC(with and arrow drawn above because it’s a ray) and BE (with and arrow drawn above because it’s a ray)
3- x=18
4- ABD=32
5- ABC=162
6- x=19
7- ABD=23
6 0
3 years ago
_+0.4=6<br><br> basic on a decimal question
erik [133]

Answer:

Your answer will be 5.6. 6.0-0.4=5.6. This is one of the fact families.

Step-by-step explanation:

Now use 5.6 in the equation. 5.6+0.4=6. This answer is correct.

3 0
3 years ago
Find the exact or approximate area of the shaded region
Lana71 [14]

Answer: 18.24 cm^2

Step-by-step explanation:

the two 135 degree angles indicate that the third angle (the one with the shaded region) is 90 degrees, which means that the shaded region is within 1/4 of the circle.

1. In order to find the area for that 1/4 region, you would have to find the area of the circle and divide it by 12

[r^2(3.14)]/4      pi is substituted by 3.14, and the radius is 8.

50.24 is the area of 1/4 of the circle.

2. Next you'd have to find the area of the triangle formed by the shaded region, and subtract the triangle from the area of the 1/4 circle.

the triangle is an isoceles triangle with two sides each valuing at 8cm, so to find the area of the triangle you would multiply the two sides (or square them since they're the same number) and divide by two

this means the triangle is 32 cm^2

3. we can subtract the 1/4 circle (50.24) by the area of the triangle (32) to get the area of the shaded region, which is <u>18.24 cm^2</u>

7 0
3 years ago
Point O is at a distance named K from the line D. What should be the value of K if we want to find three points that their dista
Keith_Richards [23]

Answer:

the answer is number 3)2

Step-by-step explanation:

subtract the both

7 0
3 years ago
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