Q: How do you write the sum of a number and 2 is 10 in equation?

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x + 10

x + (13-10)

Let's call the number "x". The given equation can be translated as: x * 3 = 2 * (x + 5). Simplifying the equation, we have: 3x = 2x + 10. Subtracting 2x from both sides gives: x = 10. So the solution to the equation is x = 10.

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class Sum_Of_Digits { public static void printSumandnoofdigits(int n) { int temp = n; int count = 0; int sum = 0; while ( n > 0 ) { sum = sum + n % 10; n = n / 10; count ++; } System.out.println("The number is..." + temp ); System.out.println("The sum of digits is..." + sum); System.out.println("The number of digits is..." + count); } }

Related questions

x + 10

The sum of 10 plus -6 is equal to 4. The equation would read 10+-6=4.

3 + 4 +10 = 17 is a number sentence that has a sum of 17. 4 + 13 = 17 is also a number sentence that has a sum of 17. Here are some more number sentences: 10 + 10 - 3 = 17 21 - 4 = 17 There are many ways to write a number sentence that has a sum of 17. What other ways can you think of?

the product of 8 and the sum of 10 and -7

Sum = 0 For N = 1 to 10 Sum = Sum + 2*N Next N Print Sum

0.6

x + (13-10)

Let's call the number "x". The given equation can be translated as: x * 3 = 2 * (x + 5). Simplifying the equation, we have: 3x = 2x + 10. Subtracting 2x from both sides gives: x = 10. So the solution to the equation is x = 10.

x+10>13 it wouldn't be x+10-13 ??

/*Reverse the digits of a number,Gankidi*/ #include<stdio.h> void main(void) { int num,sum=0,i,rem; printf("Enter a number"); scanf("%d",&num); while(num>0) { rem=num%10; num=num/10; sum=sum+(rem*10); } printf("the reverse number is %d",sum); }

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