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maksim [4K]
3 years ago
10

Find the limit, if it exists. (if an answer does not exist, enter dne.) lim (x, y) → (0, 0) x4 − 20y2 x2 + 10y2

Mathematics
1 answer:
Firdavs [7]3 years ago
5 0
Traveling along the x-axis, we have

\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^4-20y^2}{x^2+10y^2}=\lim_{x\to0}\frac{x^4}{x^2}=\lim_{x\to0}x^2=0

On the other hand, along the y-axis we get

\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^4-20y^2}{x^2+10y^2}=\lim_{y\to0}\frac{-20y^2}{10y^2}=\lim_{y\to0}(-2)=-2

Therefore the limit doesn't exist.
You might be interested in
Find the length of the diagonal of a square with perimeter of 24
julsineya [31]

Answer:

d = 6sqrt(2) or 8.4853

Step-by-step explanation:

<u><em>Formula</em></u>

P = 4*s

s^2 + s^2 = d^2 where d is the diagonal and s is the side.

<u><em>Givens</em></u>

P = 24

<u><em>Solution</em></u>

P = 4s         Substitute for s

24 = 4*s      Divide by 4

24/4 = s

s = 6

================

d^2 = s^2 + s^2

d^2 = 6^2 + 6^2

d^2 = 36 + 26

d^2 = 72

d = sqrt(72)

Factors of 72

72: 6 * 6 * 2

<em><u>Rule</u></em>: Every pair of = factors allows you to take one of them outside the sqrt sign and throw the other a way. If there are no pairs, whatever you started with stays under the root sign.

sqrt(6*6*2) = 6sqrt(2)

  • The diagonal length is either
  • d = 6*sqrt(2) or
  • d = 8.4853

8 0
4 years ago
Is parallel to 3x-5y=7 and passes through (0,-6)
oksian1 [2.3K]
Yes yes it isssssssssss
5 0
3 years ago
Math please please please help
kondaur [170]

y=3|3+2|

Hope this helps!

-Payshence

8 0
3 years ago
I need the answer asap pls! :)
AfilCa [17]

Answer:

18m

Step-by-step explanation:

Since the given traingles are similar

We get

\frac{JK}{MN} =  \frac{KL}{ML}

\frac{JK}{42}  =  \frac{21}{49}

JK =  \frac{21 \times 42}{49}

JK =  \frac{3 \times 7 \times 6 \times 7}{7 \times 7}

JK = 3 \times 6 = 18

JK = 18

Hence the distance across river is 18m

7 0
2 years ago
Please help me on this!
sashaice [31]

How does the $100 gift card affect the measure of center of the data?

It increases the mean value of the prizes.

It decreases the mean value of the prizes.

It increases the median value of the prizes.

It decreases the median value of the prizes.

8 0
4 years ago
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