10FallHwk01_Soln

10FallHwk01_Soln - Math 2413 Homework Set 1 Solution 10...

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Math 2413 Homework Set 1 – Solution 10 Points 4. (2 pts) From our calculator we get : ( ) ( ) 1 22 4 9 49 si n si n 0.2241 yy - = -& = - =- . From a quick sketch of a unit circle we can see that a positive angle corresponding to this is 2 0.224 1 6.0591 p -= . You don’t need to use this but I will for these solutions. We can also see from the unit circle that the second angle will be 0.224 1 3.3657 += . Now, all solutions are, 4 4 13.462 88 3.365 72 0 , 1 , 2, 24.236 48 6.059 12 y y yn n n n =+ & = –– K Plugging in values of n gives the following solutions. 0 : 13.4628 ny == 2 0 O R 24.2364 1 : 38.595 5 O R 49.3691 y n <= = So, we have the above three solutions. 6. (2 pts) Simplify the equation and don’t forget you can’t cancel terms from both sides of the equation when solving unless you KNOW that they won’t be zero. ( ) 1 0 2 1 1 1 1 0 21 8 1 0 3 1 0 8 3 8 30 x xx x x x - -- + = +& - e ee From this we can see that we have two solutions : 0 x = and () ( ) 1 1 1 8 1 3 3 2 13 8 3 0 1 0 2 1 l n 1 0 l n 0.4295 - = =fi- = fi=-= 7. (2 pts) First we need to combine the logarithms and then exponentiate both sides.
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This note was uploaded on 11/13/2010 for the course MA 141 taught by Professor Wears during the Spring '07 term at N.C. State.

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10FallHwk01_Soln - Math 2413 Homework Set 1 Solution 10...

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