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3 years ago

11

Assume the availability of class named DateManager that provides a static method, printTodaysDate, that accepts no arguments and

returns no value. Write a statement that calls printTodaysDate.

Computers and Technology

1 answer:

Ira Lisetskai [31]3 years ago

8 0

Answer:

Hi!

The answer is:

<em> </em> DateManager dm = new DateManager(); <em>// Instanciate.</em>

dm.printTodaysDate; <em>// Calls the method.</em>

Explanation:

If you are working on object-oriented programming, then:

First, you have to instanciate the class DataManger.

Then, you can call printTodaysDate on the instance dm of DataManager.

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Decrease in error CAD software makes use of some of the best tools, the percentage of error that occurred because of manual designing is significantly reduced.

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3 years ago

How can i clear the CMOS

motikmotik

Explanation:

Steps to clear CMOS using the battery method

Turn off all peripheral devices connected to the computer.

Disconnect the power cord from the AC power source.

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3 years ago

Find the inverse function of f(x)= 1+squareroot of 1+2x

Svetllana [295]

Answer:

Therefore the inverse function of $f(x)=1+\sqrt{1+2x}$ is $\frac{x^2-2x}{2}$

Explanation:

We need to find the inverse of function $f(x)=1+\sqrt{1+2x}$

Function Inverse definition :

$\mathrm{If\:a\:function\:f\left(x\right)\:is\:mapping\:x\:to\:y,\:then\:the\:inverse\:functionof\:f\left(x\right)\:maps\:y\:back\:to\:x.}$

$y=1+\sqrt{1+2x}$ $\mathrm{Interchange\:the\:variables}\:x\:\mathrm{and}\:y$

$x=1+\sqrt{1+2y}$

$\mathrm{Solve}\:x=1+\sqrt{1+2y}\:\mathrm{for}\:y$

$\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}$

$1+\sqrt{1+2y}-1=x-1$

Simplify

$\sqrt{1+2y}=x-1$

$\mathrm{Square\:both\:sides}$

$\left(\sqrt{1+2y}\right)^2=\left(x-1\right)^2$

$\mathrm{Expand\:}\left(\sqrt{1+2y}\right)^2:\quad 1+2y$

$\mathrm{Expand\:}\left(x-1\right)^2:\quad x^2-2x+1$

$1+2y=x^2-2x+1$

$\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}$

$1+2y-1=x^2-2x+1-1$

$\mathrm{Simplify}$

$2y=x^2-2x$

$\mathrm{Divide\:both\:sides\:by\:}2$

$\frac{2y}{2}=\frac{x^2}{2}-\frac{2x}{2}$

$\mathrm{Simplify}$

$y=\frac{x^2-2x}{2}$

Therefore the inverse function of $f(x)=1+\sqrt{1+2x}$ is $\frac{x^2-2x}{2}$

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oksian1 [2.3K]

Answer:

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Explanation:

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