3j-13=11+2j 3j-13-2j = 11+2j-2 j-13 = 11 j-13 +13 = 11 + 13 = 24 so j = 24
Julia is 3. Kate is 10. Algebra: J + (2J +4) = 13 3J +4 = 13 3J = 9 J = 3
Without information about j, the only possible answer is 3j + 15
2j+1/j+5 = 1 Multiply all terms by j+5 to eliminate the fraction: 2j+1 = j+5 2j-j = 5-1 j = 4
It is also x = 5 8zx + 5 = 2j + 3 → 8zx + 2 = 2j → 4zx + 1 = j Which is the first equation, which has a solution x = 5.
2 + 5j
3j-13=11+2j 3j-13-2j = 11+2j-2 j-13 = 11 j-13 +13 = 11 + 13 = 24 so j = 24
36.64
3i+2j)*(2i-3j)=0 bcz dot product is zero.
A = 3J; A - 5 = 5(j -5) Substitute for A: 3J - 5 = 5J - 25 Subtract 3J from each side: -5 = 2J - 25 Add 5 to each side: 0 = 2J - 20 Add 20 to each side 20 = 2J Divide each side by 2: J = 10 so Alison is now 30. (5 yrs ago they were 25 & 5)
3j + j = 4j
3-2j.
the S I unit of inductance can be written as
N = 2JD = 3J(3J + 2J + J)/3 = 6464*3 = 6J192 = 6J32 = JimNill = 2 * 32 = 64Dave = 3 * 32 = 96
Julia is 3. Kate is 10. Algebra: J + (2J +4) = 13 3J +4 = 13 3J = 9 J = 3
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Without information about j, the only possible answer is 3j + 15