Question about Mathsoft Computers & Internet

What is the largest number that will go into both 100 and 70

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7 X 2 X 5 = 70 2 X 5 X 2 X 5 = 100

7 X 10 = 70 10 X 10 = 100

10 is the largest common factor

Posted on Nov 20, 2013

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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

The worlds largest desert in actually the continent of Antarctica. It receives less than 2 inches of precipitation per year. I believe the answer you are looking for is the Sahara desert in Africa it is the largest nonpolar desert in the world. Hope the helps.

Dec 13, 2013 | Edu-Science Optics

Contact Andrews Electronics in California to see if available or they know a source---you can also try to GOOGLE the model number with "replacement screen".

Andrews is the largest OEM parts dealer in the USA and has both a 800 number and a site online.

Andrews is the largest OEM parts dealer in the USA and has both a 800 number and a site online.

May 17, 2012 | RCA HD56W66 56" HDTV

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

need to know the motherboard brand and model number.

Feb 21, 2011 | Computers & Internet

The lens aperture ring will need to be set to the largest number, and locked to that point. This allows the camera to control the aperture as it needs. If you have changed lenses to something other than what came with the camera this could also be the problem. Again make sure the aperture is set on the largest number and the problem should go away.

Jan 24, 2010 | Nikon N50 35mm SLR Camera

Alt shoud be putting out 14 to 15.2 volts.

Jul 02, 2009 | 2002 Hyundai Santa Fe

#include<stdio.h>

#include<conio.h>

void main()

{

int num,i=1,j,k;

clrscr();

printf("\nEnter a number:");

scanf("%d",&num);

while(i<=num)

{

k=0;

if(num%i==0)

{

j=1;

while(j<=i)

{

if(i%j==0)

k++; j++;

}

if(k==2)

printf("\n%d is a prime factor",i);

}

i++;

}

getch();

}

#include<conio.h>

void main()

{

int num,i=1,j,k;

clrscr();

printf("\nEnter a number:");

scanf("%d",&num);

while(i<=num)

{

k=0;

if(num%i==0)

{

j=1;

while(j<=i)

{

if(i%j==0)

k++; j++;

}

if(k==2)

printf("\n%d is a prime factor",i);

}

i++;

}

getch();

}

May 16, 2009 | Computers & Internet

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

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