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• h'(t) = -32t + 45 Set that to zero: -32t + 45 = 0 32t = 45 t = 45/32 t = 1.40625 h(1.40625) = -16(1.40625)^2 + 45(1.40625) + 40 h(1.40625) = -16(1.9775390625) + 63.28125 + 40 h(1.40625) = -31.640625 + 63.28125 + 40 h(1.40625) = 71.640625 T…

  in a few algebra problems!? PLEASE help? Comment by jeff-aaron June 2021
• h'(t) = -32t + 45 Set that to zero: -32t + 45 = 0 32t = 45 t = 45/32 t = 1.40625 h(1.40625) = -16(1.40625)^2 + 45(1.40625) + 40 h(1.40625) = -16(1.9775390625) + 63.28125 + 40 h(1.40625) = -31.640625 + 63.28125 + 40 h(1.40625) = 71.640625 T…

  in a few algebra problems!? PLEASE help? Comment by jeff-aaron June 2021
• y = (4x/1) + x^3 = 4x + x^3 The point (1,2) is not on that curve. Or did you mean: y = (4x) / (1 + x^3) Let a = 4x, so a' = 4 Let b = 1 + x^3, so b' = 3x^2 So y = a/b, so we use the Quotient Rule: y' = (a'b - ab') / (b^2) y' = ((4)(1 + x^3) …

  in Calculus problem...please help? Comment by jeff-aaron June 2021
• y = (4x/1) + x^3 = 4x + x^3 The point (1,2) is not on that curve. Or did you mean: y = (4x) / (1 + x^3) Let a = 4x, so a' = 4 Let b = 1 + x^3, so b' = 3x^2 So y = a/b, so we use the Quotient Rule: y' = (a'b - ab') / (b^2) y' = ((4)(1 + x^3) …

  in Calculus problem...please help? Comment by jeff-aaron June 2021
• Yes

  in true or false.............? Comment by jeff-aaron June 2021
• ( 6- √6) (3√2 - 2√3) (4√2 + 3√3) / 6 = (18√2 - 12√3 - 3√12 + 2√18)(4√2 + 3√3) / 6 = (18√2 - 12√3 - 6√3 + 6√2)(4√2 + 3√3) / 6 = (24√2 - 18√3)(4√2 + 3√3) / 6 = (4√2 - 3√3)(4√2 + 3√3) = 32 + 12√6…

  in Algebra problem - please help? Comment by jeff-aaron May 2021
• 5/sqrt(3) = (5*sqrt(3)) / (sqrt(3) * sqrt(3)) = (5*sqrt(3)) / (sqrt(3*3)) = (5*sqrt(3)) / (sqrt(9)) = (5*sqrt(3)) / 3 (5/sqrt(3))^2 = (5^2) / ((sqrt(3))^2) = 25 / 3

  in another algebra problem. =)? Comment by jeff-aaron May 2021
• 5/sqrt(3) = (5*sqrt(3)) / (sqrt(3) * sqrt(3)) = (5*sqrt(3)) / (sqrt(3*3)) = (5*sqrt(3)) / (sqrt(9)) = (5*sqrt(3)) / 3 (5/sqrt(3))^2 = (5^2) / ((sqrt(3))^2) = 25 / 3

  in another algebra problem. =)? Comment by jeff-aaron May 2021
• Keep dividing by prime numbers until you get a prime number. 612 / 2 = 306 306 / 2 = 153 153 / 3 = 51 51 / 3 = 17 Answer: 2 * 2 * 3 * 3 * 17

  in Factor into prime factors? Comment by jeff-aaron May 2021
• Keep dividing by prime numbers until you get a prime number. 612 / 2 = 306 306 / 2 = 153 153 / 3 = 51 51 / 3 = 17 Answer: 2 * 2 * 3 * 3 * 17

  in Factor into prime factors? Comment by jeff-aaron May 2021
• 5/sqrt(3) = (5*sqrt(3)) / (sqrt(3) * sqrt(3)) = (5*sqrt(3)) / (sqrt(3*3)) = (5*sqrt(3)) / (sqrt(9)) = (5*sqrt(3)) / 3 (5/sqrt(3))^2 = (5^2) / ((sqrt(3))^2) = 25 / 3

  in another algebra problem. =)? Comment by jeff-aaron May 2021
• 5/sqrt(3) = (5*sqrt(3)) / (sqrt(3) * sqrt(3)) = (5*sqrt(3)) / (sqrt(3*3)) = (5*sqrt(3)) / (sqrt(9)) = (5*sqrt(3)) / 3 (5/sqrt(3))^2 = (5^2) / ((sqrt(3))^2) = 25 / 3

  in another algebra problem. =)? Comment by jeff-aaron May 2021
• ( 6- √6) (3√2 - 2√3) (4√2 + 3√3) / 6 = (18√2 - 12√3 - 3√12 + 2√18)(4√2 + 3√3) / 6 = (18√2 - 12√3 - 6√3 + 6√2)(4√2 + 3√3) / 6 = (24√2 - 18√3)(4√2 + 3√3) / 6 = (4√2 - 3√3)(4√2 + 3√3) = 32 + 12√6…

  in Algebra problem - please help? Comment by jeff-aaron May 2021
• ( 6- √6) (3√2 - 2√3) (4√2 + 3√3) / 6 = (18√2 - 12√3 - 3√12 + 2√18)(4√2 + 3√3) / 6 = (18√2 - 12√3 - 6√3 + 6√2)(4√2 + 3√3) / 6 = (24√2 - 18√3)(4√2 + 3√3) / 6 = (4√2 - 3√3)(4√2 + 3√3) = 32 + 12√6…

  in Algebra problem - please help? Comment by jeff-aaron May 2021
• ( 6- √6) (3√2 - 2√3) (4√2 + 3√3) / 6 = (18√2 - 12√3 - 3√12 + 2√18)(4√2 + 3√3) / 6 = (18√2 - 12√3 - 6√3 + 6√2)(4√2 + 3√3) / 6 = (24√2 - 18√3)(4√2 + 3√3) / 6 = (4√2 - 3√3)(4√2 + 3√3) = 32 + 12√6…

  in Algebra problem - please help? Comment by jeff-aaron May 2021
• ( 6- √6) (3√2 - 2√3) (4√2 + 3√3) / 6 = (18√2 - 12√3 - 3√12 + 2√18)(4√2 + 3√3) / 6 = (18√2 - 12√3 - 6√3 + 6√2)(4√2 + 3√3) / 6 = (24√2 - 18√3)(4√2 + 3√3) / 6 = (4√2 - 3√3)(4√2 + 3√3) = 32 + 12√6…

  in Algebra problem - please help? Comment by jeff-aaron May 2021
• Yes

  in true or false.............? Comment by jeff-aaron May 2021
• y = (4x/1) + x^3 = 4x + x^3 The point (1,2) is not on that curve. Or did you mean: y = (4x) / (1 + x^3) Let a = 4x, so a' = 4 Let b = 1 + x^3, so b' = 3x^2 So y = a/b, so we use the Quotient Rule: y' = (a'b - ab') / (b^2) y' = ((4)(1 + x^3) …

  in Calculus problem...please help? Comment by jeff-aaron May 2021