Tutor profile: James G.
Remember using properties of imaginary number we know that i = sqrt(-1) We also know that because of this i^2 = -1 and i^4 = 1 Let's see if we can break down that power of 110 down to multiples of 2 or 4. Properties of exponents allows us to break down this problem. i^108(i^2) = i^110 (property of exponents) i^4 (i^27)(i^2) (4 x 27 = 108) (i^4)^27 (i^2) (property of exponents) ((1)^27) (-1) = -1 (i^2 = -1, i^4 = 1)
Assume there is a right triangle. Side 1 = 4 Side 2 = x Hypotenuse = 6 Use the Pythagorean theorem to solve for side x.
Remember that the Pythagorean theorem states that: a^2 + b^2 = c^2, where a and b are sides of a right triangle and c is the hypotenuse. Plug our side and hypotenuse values in the formula. 4^2 + x^2 = 6^2 (Solve for x and this will give us the missing side) 16 + x^2 = 36 (evaluate) 16 + x^2 = 36 (Subtract 16 from both sides isolate the x variable) -16 -16 x^2 = 20 x = Sqrt(20) (take the square root of both sides to solve for x) x = 2(sqrt(5)) (simplify)
Solve the following system of equations using the substitution method: -3x+y=15 5x+3y+31
The first step to take when using the substitution method is to take one of the equations and solve for y (isolate "y" on one side of the equation. For this problem the first equation will be easiest to do this since there is no coefficient before the "y" variable. -3x + y = 15. To solve for y add 3x to both sides of the equation: -3x + y = 15 +3x. +3x = y = 3x + 15 The next step is to substitute (3x+15) for the y variable in the second equation. That will look like this: 5x + 3y = 31 5x + 3(3x +15) = 31 Now you solve for the x: 1: 5x + 9x +45 = 31 (Distribute the 3 across 3x +15) 2: 14x + 45 = 31. (Combine like terms) 3: 14x + 45 = 31 - 45 - 45 = 14x = -14 (subtract 45 from both sides of the equation to isolate the x variable) 4: 14x/14 = -14/14 (Divide both sides by 14 to solve for x) 5: x = -1 Once we have solved for x plug the -1 for x into the first equation and solve for y. 1: -3(-1) + y = 15 (Plug in -1 for x) 2: 3 + y = 15 (Evaluate) 3: 3 + y = 15 (Subtract 3 from both sides to solve for y) -3 -3 y = 12 At this point the system of equations has been solved x = -1, y = 12
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