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sergij07
1 month ago
12

Use a table of numerical values of f(x,y) for (x,y) near the origin to make a conjecture about the value of the limit of f(x,y)

as (x,y) → (0,0) . (if the limit does not seem to exist, enter dne.) f(x, y) = x y x2 + 2 y2

Mathematics
2 answers:
tester [12.3K]1 month ago
8 0
It appears the limit to compute is

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

Taking an arbitrary line that passes through the origin y=mx, we can rewrite the expression as

\displaystyle\lim_{x\to0}\frac{mx^2}{x^2+2m^2x^2}=\lim_{x\to0}\frac m{1+2m^2}=\frac m{1+2m^2}

The limit's value is determined by the slope m selected for the line, signifying that the limit is not independent of the path and therefore does not exist.
Svet_ta [12.7K]1 month ago
3 0

Answer:

As (x,y) approaches (0,0), the limit of f(x,y) is equal to 0.

Step-by-step explanation:

The function is defined as f(x, y) = x*y/(x^2 + 2*y^2).

The corresponding data is illustrated in the attached graphic.

When y is fixed, varying x causes f(x,y) to switch between positive and negative values, and the same holds true when x is fixed and y varies. This suggests that as (x,y) approaches (0,0), the limit can be conjectured to be 0.

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Mookie Betts of the Boston Red Sox had the highest batting average for the 2018 Major League Baseball season. His average was 0.
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Answer:

Binomial probability, with n = 5, p = 0.352

Step-by-step explanation:

Each time Mookie Betts stepped up to bat, there were only two possible results: either he achieved a base hit or he didn't. The odds of hitting during each at-bat remain unaffected by prior attempts. Therefore, the binomial probability distribution applies to tackle this problem.

Binomial probability distribution

The chance of getting exactly x successes over n trials, considering p probability.

His batting average stood at 0.352.

This indicates that p = 0.352

Let's assume he bats five times in tonight's matchup against the Yankees.

This means that n = 5

a. This showcases what type of probability

Binomial probability, with n = 5, p = 0.352

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