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Korvikt [17]
2 years ago
5

Which statement is true with respect to java bytecode is platform independent

Computers and Technology
1 answer:
Rom4ik [11]2 years ago
4 0

Answer:

e with respect to java bytecode is platform indepee with respect to java bytecode is platform indepee with respect to java bytecode is platform indepee with respect to java bytecode is platform indepee with respect to java bytecode is platform indepee with respect to java bytecode is platform indepe

Explanation:

e with respect to java bytecode is platform indepee with respect to java bytecode is platform indepee with respect to java bytecode is platform indepee with respect to java bytecode is platform indepee with respect to java bytecode is platform indepee with respect to java bytecode is platform indepee with respect to java bytecode is platform indepee with respect to java bytecode is platform indepee with respect to java bytecode is platform indepee with respect to java bytecode is platform indepee with respect to java bytecode is platform indepee with respect to java bytecode is platform indepee with respect to java bytecode is platform indepee with respect to java bytecode is platform indepee with respect to java bytecode is platform indepe

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A continuous and differentiable function f(x) with the following properties: f(x) is decreasing at x=−5 f(x) has a local minimum
butalik [34]

The continuous and differentiable function where f(x) is decreasing at x = −5 f(x) has a local minimum at x = −2 f(x) has a local maximum at x = 2 is given as: y = 9x - (1/3)x³ + 3.

<h3>What is a continuous and differentiable function?</h3>

The continuous function differs from the differentiable function in that the curve obtained is a single unbroken curve in the continuous function.

In contrast, if a function has a derivative, it is said to be differentiable.

<h3>What is the solution to the problem above?</h3>

It is important to note that a function is differentiable when x is set to a if the function is continuous when x = a.

Given the parameters, we state that

f'(5) < 0; and

x = -5

The local minimum is given as:
x = -3;

the local maximum is given as

x = 3

Thus, x = -3 ; alternatively,

x = 3.  With this scenario, we can equate both to zero.

Hence,

x + 3 = 0;

3-x = 0.

To get y' we must multiply both equations to get:

y' = (3-x)(x + 3)

y'   = 3x + 9 - x² - 3x

Collect like terms to derive:

y' = 3x - 3x + 9 - x²; thus

y' = 9-x²

When y' is integrated, the result is

y = 9x - (x³/3) + c

Recall that

F (-5) < 0

This means that:

9 x -5 - (-5³/3) + c < 0
⇒ -45 + 125/3 + c <0
⇒ -10/3 + c < 0

Collecting like terms we have:
c < 10/3; and

c < 3.33


Substituting C into

f(x) = 9x - x³/3 + c; we have

f(x) = 9x - x³/3 + 3, which is the same as  y = 9x - (1/3)x³ + 3.

Learn more about differentiable functions at:
brainly.com/question/15047295
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2 years ago
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(But why would you write a rude comment in the first place to begin with?)
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Answer:

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Explanation:    

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Cheers!

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