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• You need to sum the first 18 terms in the series(That's what the S(18) means). Note that in this case, this is an arithmetic series, so the sum would actually be: Sum = (Total terms)/2 *(First term + 18th term) = (18)/2 * (5 + 3(18) + 2) = 9 *…

  in Math problem?!? Help ASAP? Comment by mecheng2030 June 2021
• Let the other two terms be x and y. It then follows that: x + y = 2.5 and 1/x = y 1/x + x = 2.5 => 1 + x² = 5x/2 => 2 + 2x² = 5x 2x² - 5x + 2 = 0 2x² - 4x - x + 2 = 0 (2x - 1)(x - 2) = 0 x = 1/2, 2 y = 2, 1/2 So th…

  in Arithmetic progress? (A.P)? Comment by mecheng2030 June 2021
• Let the other two terms be x and y. It then follows that: x + y = 2.5 and 1/x = y 1/x + x = 2.5 => 1 + x² = 5x/2 => 2 + 2x² = 5x 2x² - 5x + 2 = 0 2x² - 4x - x + 2 = 0 (2x - 1)(x - 2) = 0 x = 1/2, 2 y = 2, 1/2 So th…

  in Arithmetic progress? (A.P)? Comment by mecheng2030 June 2021
• Equation of perpendicular line: y - 9 = -1/6 * (x - 3) y = -x/6 + 1/2 + 9 => y = -x/6 + 19/2 -x/6 + 19/2 = x^2 -x + 57 = 6x^2 6x^2 + x - 57 = 0 x = (-1 +/- √(1 - 4(6)(-57))/12 = (-1 +/- 37)/12 So the points are (3, 9)(Obvio…

  in Help with parabola problem? Comment by mecheng2030 June 2021
• Equation of perpendicular line: y - 9 = -1/6 * (x - 3) y = -x/6 + 1/2 + 9 => y = -x/6 + 19/2 -x/6 + 19/2 = x^2 -x + 57 = 6x^2 6x^2 + x - 57 = 0 x = (-1 +/- √(1 - 4(6)(-57))/12 = (-1 +/- 37)/12 So the points are (3, 9)(Obvio…

  in Help with parabola problem? Comment by mecheng2030 June 2021
• f(x) = -x^2 + bx - 12 Note that the vertex form of a parabola is y = a(x - h)^2 + k, therefore from completing the square: f(x) = -(x^2 - bx + 12) f(x) = -((x - b/2)^2 + 12 - (b^2)/4) k = (b^2)/4 - 12 = 109 b^2/4 = 121 b^2 = 121*4 b = +/- sqr…

  in Pre-Calculus pls explain how to do!!? Comment by mecheng2030 June 2021
• f(x) = -x^2 + bx - 12 Note that the vertex form of a parabola is y = a(x - h)^2 + k, therefore from completing the square: f(x) = -(x^2 - bx + 12) f(x) = -((x - b/2)^2 + 12 - (b^2)/4) k = (b^2)/4 - 12 = 109 b^2/4 = 121 b^2 = 121*4 b = +/- sqr…

  in Pre-Calculus pls explain how to do!!? Comment by mecheng2030 June 2021
• e^(2x) + 2 + e^(-2x) = e^(2x) + 2 + 1/(e^(2x)) = (e^(4x) + 2e^(2x) + 1)/e^(2x) = ((e^(2x) + 1))^2/e^(2x)

  in Factor into two groups? e^(2x)+2+e^(-2x)? Comment by mecheng2030 June 2021
• u = arcsin(2x) sin(u) = 2x cos(u) = cos(arctan(2x)) = √(1 - 4x²)

  in How do you evaluate cos(arcsin(2x))? Comment by mecheng2030 June 2021
• (a.) 70,000 = Ae^(110k) 140,000 = Ae^(125k) 2 = e^(15k) => k = ln(2)/15 (b.) A = 434.05 p(0) = A = 434 approx. (c.) p(240) = 434.05*(2^16) = 28,445,901

  in calc problem help!!!? Comment by mecheng2030 June 2021
• (a.) 70,000 = Ae^(110k) 140,000 = Ae^(125k) 2 = e^(15k) => k = ln(2)/15 (b.) A = 434.05 p(0) = A = 434 approx. (c.) p(240) = 434.05*(2^16) = 28,445,901

  in calc problem help!!!? Comment by mecheng2030 June 2021
• z = x + yi => |z| = √(x^2 + y^2) = √((Re(z))^2 + (Im(z))^2) |z - 4|/|z - 8| = 1 (x - 4)^2 + y^2= (x - 8)^2 + y^2 x^2 - 8x + 16 = x^2 - 16x + 64 8x = 48 => x = 6 = Re(z).

  in Complex number problem? Comment by mecheng2030 May 2021
• x^2 + 18x - 65 (x + 9)^2 - 146

  in how do you factor x^2+18x -65? Comment by mecheng2030 May 2021
• You must understand the concept behind the equal sign. You're right in the fact that 1/cos^2(x) = sec^2(x) because that's just squaring both sides of 1/cos(x) = sec(x). Remember that whatever you do to one side of the equation, you must do to the ot…

  in if 1/cos x = sec x does 1/cos^2 x = sec^2 x ? Comment by mecheng2030 May 2021
• ∫sin(x)*cos(cos(x)) dx u = cos(x) -du = sin(x) dx -∫cos(u) du = -sin(u) + C = -sin(cos(x)) + C

  in Integral of sinx cos(cosx) dx? Comment by mecheng2030 May 2021
• ∫sin(x)*cos(cos(x)) dx u = cos(x) -du = sin(x) dx -∫cos(u) du = -sin(u) + C = -sin(cos(x)) + C

  in Integral of sinx cos(cosx) dx? Comment by mecheng2030 May 2021
• You must understand the concept behind the equal sign. You're right in the fact that 1/cos^2(x) = sec^2(x) because that's just squaring both sides of 1/cos(x) = sec(x). Remember that whatever you do to one side of the equation, you must do to the ot…

  in if 1/cos x = sec x does 1/cos^2 x = sec^2 x ? Comment by mecheng2030 May 2021
• x^2 + 18x - 65 (x + 9)^2 - 146

  in how do you factor x^2+18x -65? Comment by mecheng2030 May 2021
• x^2 + 18x - 65 (x + 9)^2 - 146

  in how do you factor x^2+18x -65? Comment by mecheng2030 May 2021
• z = x + yi => |z| = √(x^2 + y^2) = √((Re(z))^2 + (Im(z))^2) |z - 4|/|z - 8| = 1 (x - 4)^2 + y^2= (x - 8)^2 + y^2 x^2 - 8x + 16 = x^2 - 16x + 64 8x = 48 => x = 6 = Re(z).

  in Complex number problem? Comment by mecheng2030 May 2021
• z = x + yi => |z| = √(x^2 + y^2) = √((Re(z))^2 + (Im(z))^2) |z - 4|/|z - 8| = 1 (x - 4)^2 + y^2= (x - 8)^2 + y^2 x^2 - 8x + 16 = x^2 - 16x + 64 8x = 48 => x = 6 = Re(z).

  in Complex number problem? Comment by mecheng2030 May 2021
• e^(2x) + 2 + e^(-2x) = e^(2x) + 2 + 1/(e^(2x)) = (e^(4x) + 2e^(2x) + 1)/e^(2x) = ((e^(2x) + 1))^2/e^(2x)

  in Factor into two groups? e^(2x)+2+e^(-2x)? Comment by mecheng2030 May 2021
• f(x) = -x^2 + bx - 12 Note that the vertex form of a parabola is y = a(x - h)^2 + k, therefore from completing the square: f(x) = -(x^2 - bx + 12) f(x) = -((x - b/2)^2 + 12 - (b^2)/4) k = (b^2)/4 - 12 = 109 b^2/4 = 121 b^2 = 121*4 b = +/- sqr…

  in Pre-Calculus pls explain how to do!!? Comment by mecheng2030 May 2021
• Let the other two terms be x and y. It then follows that: x + y = 2.5 and 1/x = y 1/x + x = 2.5 => 1 + x² = 5x/2 => 2 + 2x² = 5x 2x² - 5x + 2 = 0 2x² - 4x - x + 2 = 0 (2x - 1)(x - 2) = 0 x = 1/2, 2 y = 2, 1/2 So th…

  in Arithmetic progress? (A.P)? Comment by mecheng2030 May 2021
• You need to sum the first 18 terms in the series(That's what the S(18) means). Note that in this case, this is an arithmetic series, so the sum would actually be: Sum = (Total terms)/2 *(First term + 18th term) = (18)/2 * (5 + 3(18) + 2) = 9 *…

  in Math problem?!? Help ASAP? Comment by mecheng2030 May 2021