2 + 5j
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
3j-13=11+2j 3j-13-2j = 11+2j-2 j-13 = 11 j-13 +13 = 11 + 13 = 24 so j = 24
Re-write it properly first, in all-algebraic language: ' 7 + 8j + 7 -2j ' Now put like terms together: ' 7 + 7 +8j - 2j ' Then simplify by doing the arithmetic: ' 14 + 6j. ' And there you are!
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.
36.64
Consider prime numbers p1, p2 greater than 2. Since p1 and p2 are prime and greater than 2, they are both necessarily odd. Hence, they are of the form: p1 = 2k+1, p2 = 2j+1, where j and k are positive integers. Their sum is then: p1+p2 = (2k+1)+(2j+1) = 2k+2j+2 = 2 (k+j+1). So 2 divides their sum, hence the sum can't be prime
3i+2j)*(2i-3j)=0 bcz dot product is zero.
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all even numbers can be written in the form 2k where, k is an integer. all odd numbers can be written in the form 2j+1, where j is an integer. 2k + 2j + 1 = 2(k+j) + 1. You can see that this is in the form of an odd number. hence, the sum of an odd number and an even number is always odd. Test the theory. Pick any even and any odd number. Add them. The total will always be odd.
It is (2j/5)2
243 mph