Alan

Take the square root of the sum of the squared coefficients of i, j and k.

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As follows:

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Not quite asinus!   ln(y) is not equal to y*ln(x) + x*ln(y)

Do dy/dx = d(x^y)/dx  +  d(y^x)/dx  first, then treat the terms on the right seperately.

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Asked and answered here: http://web2.0calc.com/questions/there-are-finitely-many-primes-p-for-which-the-congruence-has-no-solutions-x-determine-the-sum-of-all-such-p#r1

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"Sally has enough money to buy either 8d wallets or 12 coin boxes. She bought 11 wallets and 3 coin boxes. How many more wallets can she buy with her remaining money?(give your answers in term of d)"

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3^300 + 3^300 + 3^300 → 3*3^300 → 3^301

3^301 = 3^x  means x = 301

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"At 7am, Alex set off at a speed of 60km/h from Town P on a journey to Town Q. At 8.15 am, Bruno set off from Town P. He took the same route as Alex and overtook him at 10.45 am. At this point, Bruno had completed 5/8 of his journey. If Bruno then increased his speed by 10km/h, how far was Alex from Town Q when Bruno reached Town Q?"

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In logic "and" means that both elements must be true for the combination to be true.

In arithmetic or algebra it can usually be replaced by "multiplied by".

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Graphs are good!! 

1. Form the cross product of u and v:

uxv = (-3i - j + k)x(-3i + 2j + 3k)

→ (-3i)x(-3i) + (-3i)x2j + (-3i)x3k 

   + (-j)x(-3i) + (-j)x2j   + (-j)x3k

   + kx(-3i)    +  kx2j     +  kx3k

→ 9ixi - 6ixj -9ixk + 3jxi -2jxj - 3jxk - 3kxi + 2kxj + 3kxk

Now for cross products we have. ixi = jxj = kxk = 0

                                                     ixj = -jxi = k

                                                     jxk = -kxj = i

                                                     kxi = -ixk = j

So: uxv → 0 - 6k + 9j -3k - 0 - 3i - 3j - 2i + 0 

      uxv → -5i + 6j - 9k

2. To get the unit vector divide each coefficient by sqrt(5^2 + 6^2 + 9^2) → sqrt(142)

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