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Enter a;

enter b;

enter c;

if a>=b and a>c

print "a is the greatest number"

then if b>=a and b>c

print "b is the greatest number"

then if a=b and a=c

print "a, b, c are equal"

if not print "c is the greatest number"

this is pseudocode.. or my drunken attempt to remember pseudocode... but I think the logic could work.

Posted on Nov 27, 2008

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Posted on Jan 02, 2017

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The largest three-digit number is 999. The largest one-digit number is 9. The product of 999 and 9 is 8991.

Oct 02, 2014 | Office Equipment & Supplies

please answer your product's id number.visual basic 6.0

Thanks.....

Thanks.....

Feb 26, 2013 | Microsoft Mlf Visual Studio Enterprise...

Write a program to find out the largest number from
a

given unordered array of 8-bit
numbers. Store the largest

number in largest variable.

** **

** 52h, 23h, 56h, 45h, 9Ah,
ABh **

** **

Aug 12, 2011 | Borland Turbo Assembler and Debugger 5.0...

The hardest part of a word problem is translating it into the equation or equations you need. This one needs just one equation.

Let N = the smallest number. Since they are consecutive, the next ones will be N+2 and N+4. (It doesn't matter that they are odd, just that each one is 2 greater than the one before.)

So, N = the smallest number and N+4 = the largest. Translate the problem (3 times the smallest, N, decreased by 7, equals twice the largest, N+4) into numbers:

3N - 7 = 2 (N+4)

Then it's simple to find N:

3N - 7 = 2N +8

3N - 2N = 8 + 7

N = 15

The three numbers then are 15, 17, 19. Check the answer by seeing if they fit the original problem. 3 X 15 = 45, 45 - 7 = 38, 2 X 19 = 38, so we got the right values.

Let N = the smallest number. Since they are consecutive, the next ones will be N+2 and N+4. (It doesn't matter that they are odd, just that each one is 2 greater than the one before.)

So, N = the smallest number and N+4 = the largest. Translate the problem (3 times the smallest, N, decreased by 7, equals twice the largest, N+4) into numbers:

3N - 7 = 2 (N+4)

Then it's simple to find N:

3N - 7 = 2N +8

3N - 2N = 8 + 7

N = 15

The three numbers then are 15, 17, 19. Check the answer by seeing if they fit the original problem. 3 X 15 = 45, 45 - 7 = 38, 2 X 19 = 38, so we got the right values.

Mar 06, 2011 | Office Equipment & Supplies

I am pretty sure I got this as a homework assignment in programming class. If I had someone give it to me then I wouldn't be even the limited programmer I am now. That said, I will help, but I will not give you the full out answer.

dia star = number of lines before star and "*"

print dia star

NOLA Tech

- The number of stars and characters per line are an odd number to make the ASCii look like a diamond. For my example, the maximum number of characters per line is five.
- Create a variable called dia_star that will hold the spaces and stars. To figure out the number of spaces that should be put into each star line, use this formula:

dia star = number of lines before star and "*"

print dia star

- --*--
- -***-
- *****

NOLA Tech

Jan 30, 2010 | Microsoft Visual Basic Enterprise Edition...

Let the three consecutive numbers be x, x+1, x+2.

Now, the smallest is 57 less than three times the largest...

=> x + 57 = 3(x + 2)

=> x + 57 = 3x + 6

=> 2x = 57 - 6

=> 2x = 51

=>** x = 25.5**

=> x, x+1, x+2 = 25.5, 26.5, 27.5

So, the three consecutive numbers are**25.5, 26.5, 27.5**.

Now, the smallest is 57 less than three times the largest...

=> x + 57 = 3(x + 2)

=> x + 57 = 3x + 6

=> 2x = 57 - 6

=> 2x = 51

=>

=> x, x+1, x+2 = 25.5, 26.5, 27.5

So, the three consecutive numbers are

May 11, 2009 | The Learning Company Achieve! Math &...

This code generates some random number to test.

#include <limits.h>

#include <stdio.h>

#include <stdlib.h>

int main()

{ findLargest();

return 0;

}

int findLargest()

{ int someNumbers[50];

int i;

// to generate some numbers for us to test

for(i=0; i<50; i++)

// generate a random number between 0-100

someNumbers[i] = rand() % 100;

// for keeping track of numbers, set as smallest possible

int largest = INT_MIN;

int largest2 = INT_MIN;

// go through each item in the array

for(i=0; i<50; i++)

{ // if bigger than our previous max, set as new max

if (someNumbers[i] > someNumbers[largest])

largest = i;

// if it's not been set as new max, and is bigger than current 2nd largest

else if(someNumbers[i] > someNumbers[largest2])

largest2 = i;

// for printing all numbers in the array

printf("%d | %d\n",i, someNumbers[i]);

}

// print largest numbers and their position in the array

printf("largest %d (pos %d).\n2nd largest %d (pos %d)",

someNumbers[largest],

largest,

someNumbers[largest2],

largest2

);

return 0;

}

#include <limits.h>

#include <stdio.h>

#include <stdlib.h>

int main()

{ findLargest();

return 0;

}

int findLargest()

{ int someNumbers[50];

int i;

// to generate some numbers for us to test

for(i=0; i<50; i++)

// generate a random number between 0-100

someNumbers[i] = rand() % 100;

// for keeping track of numbers, set as smallest possible

int largest = INT_MIN;

int largest2 = INT_MIN;

// go through each item in the array

for(i=0; i<50; i++)

{ // if bigger than our previous max, set as new max

if (someNumbers[i] > someNumbers[largest])

largest = i;

// if it's not been set as new max, and is bigger than current 2nd largest

else if(someNumbers[i] > someNumbers[largest2])

largest2 = i;

// for printing all numbers in the array

printf("%d | %d\n",i, someNumbers[i]);

}

// print largest numbers and their position in the array

printf("largest %d (pos %d).\n2nd largest %d (pos %d)",

someNumbers[largest],

largest,

someNumbers[largest2],

largest2

);

return 0;

}

May 07, 2009 | Computers & Internet

http://www.microsoft.com/visualstudio/en-us/products/teamsystem/default.mspx?pt_id=-1&WT.mc_id=B66C779D-0907-4EE1-9435-FED85F10C8E2&WT.srch=1&wt.mc_id=vspdsrch

this site has a free sample download

this site has a free sample download

Apr 16, 2009 | Microsoft Visual Basic Professional...

write the following function and call it whenever you need to evaluate a number:

Private Function IsPrime(ByVal Number As Long) IsPrime = False Dim I As Long For I = LBound(Primes) To UBound(Primes) DoEvents Call UpdateStatus(Number, Primes(I)) If (Number Mod Primes(I) = 0) Then Exit Function If (Primes(I) >= Sqr(Number)) Then Exit For Next IsPrime = True End Function

Private Function IsPrime(ByVal Number As Long) IsPrime = False Dim I As Long For I = LBound(Primes) To UBound(Primes) DoEvents Call UpdateStatus(Number, Primes(I)) If (Number Mod Primes(I) = 0) Then Exit Function If (Primes(I) >= Sqr(Number)) Then Exit For Next IsPrime = True End Function

Jan 06, 2009 | Microsoft Visual Basic 6.0 for PC

Q.2# Dim n, sum as Double

Dim i as Integer

n=100

sum=0

for i=1 to n

sum=sum+1/n

next i

Q.3# Dim sum as Double

sum=0

for i=1 to 99

if i%2 <> 0 then sum=sum+i end if

next i

Dim i as Integer

n=100

sum=0

for i=1 to n

sum=sum+1/n

next i

Q.3# Dim sum as Double

sum=0

for i=1 to 99

if i%2 <> 0 then sum=sum+i end if

next i

Sep 16, 2008 | Microsoft AccRepair DS2 2004 for Fr...

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