Hi, I need to find the absolute value of the complex number, in these examples-\n\n1). 2 + i\n2). 3i -2 \n\nI could really use and appreciate some help. Any ideas would be extremely helpful considering this is a take-home test for math =) Thanks!
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The absolute value of the complex number x + iy is defined as sqrt (x^2 + y^2).\nSo,\n1) |2 + i| = sqrt(2^2 + 1^2) = sqrt 5\n2) |3i - 2| = sqrt(3^2 + (-2)^2) = sqrt 13
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http://www.math.psu.edu/tseng/class/Math140H/Sqrti.html\n\nAnd, BTW, i (or j) is not a 'complex' operator.  It is the 'imaginary unit' ☺\n\n\nDoug
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1) yes, by definition\n2) no, example: sin x on [0,2 pi]\n3) no , example |x|, absolute value of x is continuous in 0  but not differentiable in 0.\n4) yes, since (F-G)' = f-f = 0, then F-G is a constant.\n5) yes, known theorem
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1. absolute value of any # is always positive.....so -7.5=7.5\n2. simplify, evaluate, solve....in math are pretty much the same things....in this case u only hav to bring out the 17 out of the absolute value. therefore ur ans wud b 17.\n3. devide -18/25, which is -0.72. since its in absolute value...the final ans wud b 0.72\nB. add 3+-4, which cud also b written as 3-4. the ans for that will b -4. since again its in absolute value, the final ans will b 1.
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lim ((arctan(x))-(pi/2))*(x) = (lim(arctan(x)-lim(pi/2))*(x)\nx->+infinity                            is this step ok?\n\nanyway: looks like it's 0*x, so 0
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the answer doesnt work out correctly because you have a negative under the square root of the quadratic equation.  but it can be solved using i.  then the answer would be 2.33i and -1.73i
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First of all is the radical in the numerator over just the 10 or over the entire 10y^3?\n\nThe only this you can factor out and cancel is a radical 2 on the top and bottom.  Other than that, you can't really simplify this problem.  Are you sure you wrote it down correctly?
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Let Z = 3 - 3i\n|Z| = √(3² + 3²)\n= 3√2\nThus Z = 3√2(1/√2 - 1/√2i)\nNow swwk an angle such that cosθ = 1/√2 and sinθ = -1/√2\nVlearly this is a 4th quadrabt angle\nTherefore θ = 2π - π/4\n= 7π/4\nThus Z = 3√2(cos7π/4 + i sin7π/4)\n[= 3√2e^(i7π/4)]
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I=I1+I2; I1=int{for x=pi/6 until pi/2} of x*dx = x^2/2{for x=pi/6 until pi/2}= pi^2/9;\nI2=int{for x=pi/6 until pi/2}of  25*dx/(sinx)^2; \nLet t=tg(x), then 1/(sinx)^2=1/t^2+1 and dx=(1+t^2)*dt;\nI2=int{for t=1/sqrt(3) until +inf}of 25*t^(-2)*dt = -25/t = 25*sqrt(3)
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Two complex solutions because x= -b + or - the square root of b2-4ac all divided by 2a (if that made any sense, I can't really write it) and since you'd have to have the square root of a negative number, it has to be complex.  And since it's plus or minus, you have two answers!
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find whole (integer) number a,b, and c such that a^3+b^3=c^3.\nDon't remember the answer anymore,but it should be easy - all numbers were less that 10.
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No, watch your -18i^2.\n-18*-1 = 18\nLook at your second to last step.
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I'm in my final year of university and I still cannot remember the quadratic formula. And my degree in in math. haha becoz it can be easily derived, so why waste your brain cell remembering it?\n\n\nanyway, when sqrt(b^2-4ac)<0, meaning it has no REAL root**, it means the quadratic equation has no REAL solution.\n\nWhen I say REAL, I don't mean real or fake. I mean, REAL NUMBER solution. In other words, it does not cross over at the x-axis.\n\n\n\n**It does have a solution in the complex number space though. As u might know, when u sqrt(-1) u get i, the imaginary number. but u don't need to know that yet.
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Alright, don't listen to the above post...\n\nGrade 12 here, answering questions all day for mere fun.\n\n12)\ny - 3 = -1/3 (x + 4)\n\nEXPLAINED\n\nadd 3 to both sides\ny = -1/3 (x + 4) + 3\n\nexpand 1/3(x + 4) \n= x * 1/3 + 4 * -1/3\n= x/3 + 4/3\ny = -1/3x - 4/3 + 3\n\nTo add -4/3 and 3, use 3 = 9/3\ny = -1/3x - 4/3 + 9/3\ny = -1/3x + 5/3\n\nTrue, my bad. I originally added 4/3 instread of -4/3\n\n\n16)\n(-2, -1) (4, -4)\n\nm = Dy / Dx\nm = ((-1) - (-4)) / ((-2) - (4))\nm = 3 / -6\nm = -1/2\n\ny = mx + b\n(-1) = -1/2(-2) + b\n-1 = 1 + b\n-1 - 1 = b\n-2 = b\n\ny = -1/2x - 2\n\nCHECK\ny = -1/2(4) - 2\ny = -2 - 2\ny = -4\nTherefore it works.\n\n18)\n(0, 0) (2, 3)\n\nm = Dy / Dx\nm = ((3) - (0)) / ((2) - (0))\nm = 3/2\n\nb = 0\n\ny = mx + b\ny = (3/2)x + 0\ny = (3/2)x\n\nNote, the lower post copied my answers
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That simplifies to 3[2x-1]>12\n\n[2x-1]>4\n\n2x-1>4    2x-1<-4\n\nx>5/2  or x<-3/2\n\nWhen you have absolute values, you have to look at both ends of the spectrum
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I'll write out the steps, one by one.\nGiven: 2ab/(a^2-b^2)  - b/(a-b)  + 3\n=> 2ab/(a^2 - b^2) - b(a+b)/(a^2-b^2) + 3\n=> (2ab - b(a+b))/(a^2-b^2) + 3\n=> (2ab - ab - b^2)/(a^2-b^2) + 3\n=> (ab - b^2)/(a^2-b^2) + 3\n=> (b(a-b))/(a^2-b^2) + 3\n=> b/(a+b) + 3\n=> (b + 3(a+b))/(a+b)\n=> (3a+4b)/(a+b)\nQED.\n\n\nYou are using the fact that (a^2-b^2) is the same as (a+b)(a-b)
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Ok, Victoria, if you're going to ask for help you have to write the questions correctly.  Problem 1 cannot be solved using algebra the way you wrote it, I doubt you can use complex integration or any other advanced tool to solve that one.  Unfortunately, I just spent the last half hour manipulating it before I remembered that little fact.\n\nPut parentheses in.  Had you put the parentheses in the problem would have been trivial.  Kudos to the helpers who were smarter than I in figuring out the constants were in the exponent.  It's not that hard Victoria, especially when you're asking for help.\n\nWrite the question correctly.\n\n\nPS. Irma, you forgot to multiply the 2 from 4=2^2 through the whole parenthetical statement.  The answer is x=4.\n.