Calculus
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asked by girly girl on March 22, 2018 
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1 last square root : Rationalize the denominator 5/sqrt[3]+sqrt[5]= 5*sqrt[3]sqrt[5]/ sqrt[3]sqrt5[5]*sqrt[3]sqrt[5]= 5sqrt[3]5sqrt[5]/sqrt[9]sqrt[15]+sqrt[15] sqrt[25]= 5sqrt[3]5sqrt[5]/16 = 5sqrt[3]5sqrt[5]/ 2
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simplify: square root of 5 + square root of 20  square root of 27 + square root of 147. simplify: 6/4square root of 2 sqrt 5 + sqrt 20  sqrt 27 + sqrt 147 = sqrt 5 + sqrt (4*5)  sqrt (3*9) + sqrt (49*3) = 3 sqrt 5  3 sqrt 3 +
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Evaluate sqrt7x (sqrt x7 sqrt7) Show your work. sqrt(7)*sqrt(x)sqrt(7)*7*sqrt(7) sqrt(7*x)7*sqrt(7*7) sqrt(7x)7*sqrt(7^2) x*sqrt 7x49*x ^^^ would this be my final answer?
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In these complex exponential problems, solve for x: 1)e^(i*pi) + 2e^(i*pi/4)=? 2)3+3=3i*sqrt(3)=xe^(i*pi/3) MY attempt: I'm not really sure of what they are asking. For the 1st one I used the e^ix=cos(x)+i*sin(x) and got
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Calculus  Second Order Differential Equations
Solve the initialvalue problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/ Sqrt(4^24(1)(6)))/2(1) r=(16 +/ Sqrt(8)) r=8 +/ Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2
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Find the area of the region bounded by the curves y=x^(1/2), y=x^(2), y=1 and y=3. You get: a.) 1/2(sqrt(3)) + 4/3 b.) 2(sqrt(3))  8/3 c.) 1/2(sqrt(3)  32/3 d.) 2(sqrt(3))  32/3 e.) 8/3  2(sqrt(3))
asked by Mishaka on February 29, 2012 
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Which is the exact value of the expression sqrt 32 sqrt 50 + sqrt 128 2 sqrt 7 7 sqrt 2 22 sqrt 2 2 sqrt 55
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