your equation becomes: 4 = 4x^2/(1-x)^2 multiply both sides by (1-x)^2 4(1-x)^2 = 4x^2 4(1 - 2x + x^2) = 4x^2 4 - 8x + 4x^2 = 4x^2 4=8x x = 1/2
answered by Steve yesterday http://www.jiskha.com/display.cgi?id=1368459584
Is it tan (x/2) = tanx/secx + 1 ?? tan (x/2) = tanx/(secx +1) ?? (tanx)/2 = tanx/secx + 1 ?? ---> the way you typed are we solving the equation, or is it supposed to be an identity? Can you see why brackets are essential?
Let the depth of the water be d(t) if d ' (t) = 4t+5 , where t is in hours d(t) = 2t^2 + 5t + c, where c is a constant given: when t=0, d(0) = 2.5 2.5 = 0 + 0 + c , ----> c = 2.5 d(t) = 2t^2 + 5t + 2.5 a) when t = 20 min, t = 1/3 hrs d ' (1/3) = 4(1/3) + 5 = 19/3 m/...
π/6 = 30° 2π/3 = 120° leaving 180 -120 - 30 or 30° for the third angle
So you want (95-2x) (25+3x) = 3600 2375+235x -6x^2 - 3600 = 0 6x^2 - 235x +1225 = 0 by the quadratic formula: x = 6.19 or x = 32.9 check: if x = 6, number sold is 25+18 = 43 , price of each = 95-12 = 83 income = 43(83) = $3569 , close enough to $3600 if x = 33 , number sold = ...
secAcotB-secA-2cotB+2=0 (1/cosA)(cosB/sinB) - (1/cosA)- 2(cosB/sinB) = -2 multiply each term by sinBcosA cosB - sinB - 2cosBcosA = -2cosAsinB cosB - sinB - 2cosBcosA + 2cosAsinB = 0 (cosB - sinB) - 2cosA(cosB - sinB) = 0 (cosB - sinB)(1 - 2cosA) = 0 cosB = sinB OR 1 - 2cosA = ...
Let those number of years be x in x years: Pam will be 14+x Dad will be 37+x when is 37+x = 2(14+x) 37+x = 28+2x =x = -9 x = 9 It will happen 9 years from now check: in 9 years, she will be 23 , Dad will be 46 which is twice her age.
2) I thought there was something wrong with your first typing of the question 1/B = kt + 1/A multiply each term by AB, the LCD A = ABkt + B A - B = ABkt t = (A-B)/(ABk) k = (A-b)/(ABt) from: A = ABkt + B A = B(Akt + 1) B = A/(Akt + 1)
1) ln(A) = -kt+ln(B) kt = lnB - lnA kt = ln(B/A) t = ln(B/A) / k k = ln(B/A) / t lnB = lnA + kt B = e^(lnA + kt) 2) 1/B = kt + 1/B 0 = kt t = 0 or k = 0 B is any value except B≠0 (are you sure you have no typo in 2) ?
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