1 and 7/12 = 19/12
5 and 5/6 = 35/6 = 70/12
4 and 2/3 = 14/3 = 42/12
19 + 70 = 89
89 - 42 = 47
47/12 = 3 and 11/12
3x+1+2x-2 = 64 3x+2x = 64-1+2 5x = 65 x = 13
2/12 + 2/12 + 4/12 + 9/12 + 6/12 = 23/12 or 1 and 11/12
1, 2, 4, 8, 16, 32, 64 1, 2, 3, 4, 6, 12, -1, -2, -3, -4, -6, -12
Common factors of 12 & 64 are 1, 2 and 4.
No, 12 is not a divisor of 64. These are 64's divisors: 1, 2, 4, 8, 16, 32, 64.
We can use the identity ((a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc)). Given that (a + b + c = 12) and (a^2 + b^2 + c^2 = 64), we can substitute these values into the identity: [ 12^2 = 64 + 2(ab + ac + bc). ] Calculating (12^2) gives us 144, so: [ 144 = 64 + 2(ab + ac + bc). ] Subtracting 64 from both sides gives us: [ 80 = 2(ab + ac + bc). ] Dividing by 2, we find: [ ab + ac + bc = 40. ] Thus, the value of (ab + ac + bc) is 40.
1 + 2 + 1000 + 300 + 23 + 12 = 1338
It is: 1+9+2 = 12
yes. think of 6/12 plus 1/12. that is 7/12!
12x^2+24x+12 12(x^2+2x+1) 12(x+1)^2
1+12=13+15=28+2=30+29=59.
The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24The factors of 64 are: 1, 2, 4, 8, 16, 32, 64The factors of 88 are: 1, 2, 4, 8, 11, 22, 44, 88