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Placing a question mark at the end of a statement does not make it a sensible question. Try to use a whole sentence to describe what it is that you want answered.Yes, it is possible to input a number and find whether it is positive or negative. Yes, it is possible to write an algorithm to do that. But what is it that you want?

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Q: Input a number and find whether it is positive or negative?

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The rule that determines the output number based on the input number is known as a function. For example take the function: f(x) = x+1. F is the name of our function, x is the input number, and f(x) is our output number. So if our input number is 3, our function or "rule" says to add one to it. Therefore, f(x), known as the output number, would be 4 since 3+1 = 4.

yes it is. http://www58.wolframalpha.com/input/?i=is+151+a+prime+number%3F

complex to-complex(mag, theta) { if(mag >=0) { real = mag * cos(theta); img = mag * sin(theta); return real + i * img; } else raise error; } One test is needed : mag must be a positive number! And the return value is depending of your way to deal with complex number

No. It is 3 x 1039. (Note: I got this from the Wolfram Alpha site; input: "factor 3117".

Use Wolfram|Alpha... go to the related link below, Wolfram|Alpha, and type in (is __ (number) prime) and then the program will compute that and tell you if it is prime or composite.

Related questions

draw a flow chart to input ten number and count the positive negative and zero

When the positive differential input terminal of an OP AMP is linked to the output, we have positive feedback and when the negative differential input terminal is linked to the output , we have negative feedback.

A: Any feedback that contributes to the input is positive feedback any feedback that subtract from the input is negative feedback

Positive clipper-the clipper which removes the positive half cycles of the input voltage, while the negative clipper the clipper which removes the negative half cycles of the input voltage.

Ideal op amp approximations: -no current goes into the positive or negative input of the op amp. -The open loop gain is infinite. -Voltage at positive input is the same as the negative input.

That is a function defined as: f(x) = -1 if x is negative f(x) = 0 if x is zero f(x) = 1 if x is positive In other words, a function that basically distinguishes whether the input is positive, negative, or zero.

in what languange? I assume it's TI-Basic (or something else)TI-Basic:input Aif A < 0thendisp "Negative"elseif A > 0thendisp "Positive"elseif A = 0thendisp"zero"endor PHP:$a = $_POST['input'];if ($a < 0) {echo $a . " is smaller then one, meaning it's a negative number.";} else if ($a > 0) {echo $a . " is bigger then one, meaning it's a positive number.";} else {echo $a . " is a zero"}I'm sure you can convert it to c++ or something else. It's as simple as that!

There are no real answers for a square root of a negative number. Try it on a good scientific calculator (some calculators simply square root the positive, ignoring the negative input), and it should come up with a math error.

Recall that domain is the input of a function, and the range is the output. If the value for the input (domain) represents the height of a person, it would not be reasonable to use a negative number for the input because this would indicate a negative height. That would not be realistic or negative. Some events would be reasonable or realistic if a negative number would be used as a valid time input and some events would not. For example, a negative time may represent the time before blastoff, but a negative time would not be realistic if it was used to represent the time you spent on homework last week. Only zero or a positive number would be realistic, therefore the reasonable domain would be non-negative numbers (positive numbers and zero).

#include#include#include#includeusing namespace std;string ask(string prompt){string input;coutinput)return(input);return(0);}bool ask_yn(std::string prompt){while(1){string input = ask(prompt);if( input.size() && input.size()

clamper

It will be pegged to the positive rail. The only time the output of an op amp is between the rails is when the two inputs are equal, within the offset error of the op amp. You are describing an open loop condition and, unless there is a 2V offset error, the output will be pegged; in this case to the positive rail since the positive input is greater than the negative input. Also, the answer depends on whether or not the inputs are within range. If, for instance, the op amp used +/- 5V as its power, then the 7V input would be out of range, and the op amps behavior might be unpredictable. (It might lockup, it might be destroyed, etc.)

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