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Posted on Jan 02, 2017

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I am assuming you are just making a triangle, not a right-angled triangle.

Think of these as two separate pieces of wood. The largest possible third side would be the two pieces of wood end to end, with them just a tiny tiny tiny bit less than 180 degrees, nearly forming a straight line. In this case, the third side would be just under 14 units long. So the largest whole-numbered length would be 13 units.

Good luck,

Paul

Think of these as two separate pieces of wood. The largest possible third side would be the two pieces of wood end to end, with them just a tiny tiny tiny bit less than 180 degrees, nearly forming a straight line. In this case, the third side would be just under 14 units long. So the largest whole-numbered length would be 13 units.

Good luck,

Paul

Jun 01, 2018 | Homework

how about 987

Mar 05, 2015 | Super Tutor Pre Algebra (ESDPALG)

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 preceding describes six decimal numbers:

3517

3715

5317

5713

7315

7513

The largest of these is 7513 .

3517

3715

5317

5713

7315

7513

The largest of these is 7513 .

Jun 16, 2011 | Office Equipment & Supplies

Odd numbers 1, 3, 5, 7

Uses all 4

1 is in the 10's place

7513 is the largest number that can be created

This is answer based on the provided information.

Uses all 4

1 is in the 10's place

7513 is the largest number that can be created

This is answer based on the provided information.

Jun 16, 2011 | Casio Office Equipment & Supplies

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

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

Please see the chart at the bottom of this page at Fuji

This page is for the 8100fd, which is a slightly different model, but the number of photos should be substantially similar for both models.

This page is for the 8100fd, which is a slightly different model, but the number of photos should be substantially similar for both models.

Dec 26, 2008 | Fuji FinePix S8000fd Digital Camera

Aug 21, 2018 | Cell Phones

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