If 4m=10 what is the value of 12m

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

1 year ago

I need answers right now

natali 33 [55]

Answer:

a m not with math dry buddy hope u find somebody to help pu

5 0

2 years ago

Which equation does not represent a direct variation?

tatuchka [14]

Okay I think there has been a transcription issue here because it appears to me there are two answers. However I can spot where some brackets might be missing, bear with me on that.

A direct variation, a phrase I haven't heard before, sounds a lot like a direct proportion, something I am familiar with. A direct proportion satisfies two criteria:

The gradient of the function is constant s the independent variable (x) varies

The graph passes through the origin. That is to say when x = 0, y = 0.

Looking at these graphs, two can immediately be ruled out. Clearly A and D pass through the origin, and the gradient is constant because they are linear functions, so they are direct variations.

This leaves B and C. The graph of 1/x does not have a constant gradient, so any stretch of this graph (to y = k/x for some constant k) will similarly not be direct variation. Indeed there is a special name for this function, inverse proportion/variation. It appears both B and C are inverse proportion, however if I interpret B as y = (2/5)x instead, it is actually linear.

This leaves C as the odd one out.

I hope this helps you :)

4 0

3 years ago

2×-20/3 =2x help me please

Llana [10]

Answer:

x = - 5

Step-by-step explanation:

Given

$\frac{2x-20}{3}$ = 2x ( multiply both sides by 3 to clear the fraction )

2x - 20 = 6x ( subtract 2x from both sides )

- 20 = 4x ( divide both sides by 4 )

- 5 = x , that is x = - 5

4 0

3 years ago

Suppose that F(x) = x^2 and G(x) = 2x^2-5. Which statement best compares the graph G(x) with the graph of F(x)?

notsponge [240]

I believe the correct answer from the choices listed above is option D. The graph <span>G(x) as compared to the graph of F(x) would be that the </span><span>graph of G(x) is the graph of F(x) stretched vertically and shifted 5 units down. 2 is a stretch factor and -5 is the shift downwards of the graph. Hope this answers the question.</span>

8 0

3 years ago

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