engr1 hw1

# engr1 hw1 - both sides of the equation = 2.37483 C Problem...

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Stephen Jabaut September 2 nd , 2007 Engineering 1 hw1 Problem 1. The program will solve the equations to find various values of C and i2. The given values of a,b,d, i1, and i3 are used to find the unknown values. 7.1*1.4+(-1.65) = C = 8.29 7.1/(1.4+1.65) = C = 2.32787 (7.1/1.4)*-1.65 = C = -8.36786 7.1/1.4/(-1.65)*4 = C = -12.2944 7.1*1.4 = i2 = 9 (9.94 is rounded down to 9 due to int ) 1.4/4 + 1.4 = i2 = 1 (1.75 is rounded down to 1 due to int ) The remainder of 6/4 = i2 = 4 (1.4/6)/7.1 = C = .0328638 (1.4+7.1)/7.1 + 1.4(7.1+C)*i3 = C (I predict that this wil l an answer plus the variable C because C is on

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Unformatted text preview: both sides of the equation) = 2.37483 + C Problem 2 Problem 3 Problem 4 using namespace std; #include <oistream> //the first error is right here, one can see that the oi in the “<iostream>” is reversed int main { } //this is the second error one should use () these brackets. { } these ones have a different meaning in the compiler. { double total; integer n; total = 5.2 total = 3.1 *total n = total/ 4.2 ; cout>>/*here the arrows are facing the wrong way it should be <<*/ ”n is equal to “ <<n<<endl; system( “pause” ); return 0; }...
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engr1 hw1 - both sides of the equation = 2.37483 C Problem...

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