Recent content by AN630078

  1. AN630078

    Iterative root finding for the cube root of 17

    Firstly, the cube root of 17 is 2.571281591 which is 2.57 to 3.s.f. Initially, I thought about just approaching this problem using the Newton-Raphson Method when x0=2. In which case; x^3=17 x^3-17=0 Using the Newton-Raphson iterative formula xn+1=xr-f(xn)/f’(xn) f(x)=x^3-17 f’(x)=3x^2...
  2. AN630078

    Modulus Inequalities -- Help Needed Please

    Thank you for your reply. Yes, I found found that with nonlinear inequalities. Graphs certainly have been helpful to my understanding this topic better. Although, I still find it a little confusing. Thank you for your suggestion for the alternative method of solving the second part to this...
  3. AN630078

    Modulus Inequalities -- Help Needed Please

    Thank you for your reply. For 1, yes sorry my workings should simplify -3x+2-1/x<0 by converting to a fractional form; -3xx/x+2x/x-1/x<0 -3x^2+2x-1/x<0 Multiply both sides by -1 to reverse the inequality; 3x^2-2x+1/x>0 Combine the intervals x>0 and 0<x<2/3, the solution is 0<x<2/3. Yes, I...
  4. AN630078

    Solving Reciprocal Trigonometric Equation cot^2θ+5cosecθ=4

    No do not worry, I am pleased you have used it, now I know its meaning 😁
  5. AN630078

    Solving Reciprocal Trigonometric Equation cot^2θ+5cosecθ=4

    Oh, ok I was confused by the "something". Thank you for explaining 👍
  6. AN630078

    Solving Reciprocal Trigonometric Equation cot^2θ+5cosecθ=4

    Thank you for your reply. Oh splendid, just to ask what do you mean be 9.8°sth?
  7. AN630078

    Modulus Inequalities -- Help Needed Please

    1. I think the question is asking where is the graph of |3x-2| below the graph of 1/x. To sketch the graph of y= |3x-2| draw the line of y=3x-2 and reflect the section with negative y-values in the x-axis. Alternatively, I could set 3x-2 ≥0, meaning |3x-2|=3x-2 so draw the line of y=3x-2. Then...
  8. AN630078

    Solving Reciprocal Trigonometric Equation cot^2θ+5cosecθ=4

    Thank you for your reply, so the solutions in the range 0 °<θ<360 ° would be θ=189.8 °and θ=350.16°?
  9. AN630078

    Solving Reciprocal Trigonometric Equation cot^2θ+5cosecθ=4

    cot^2θ+5cosecθ=4 cot^2θ+5cosecθ-4=0 cosec^2θ+5cosec-4-1=0 cosec^2θ+5cosec-5=0 Let u=cosecθ u^2+5u-5=0 Solve using the quadratic formula; u=(-5± 3√5)/2 u=(-5+ 3√5)/2=0.8541... Substitute cosecθ=u Therefore, cosecθ=0.8541 1/sinθ=0.8541 sinθ=1/0.8541=1.170... which is not true since sin x cannot be...
  10. AN630078

    Composite and Inverse Functions Problem

    Thank you very much for your informative reply. Right, I may add the domain and range of each function for f(x) and g(x) in part a. Really, I used the notation fg(x) and gf(x) because this was what the question specified and additionally what I had been taught in my textbook, which explained...
  11. AN630078

    Composite and Inverse Functions Problem

    1. a. fg(x)=2(1/2(x-1))+1 fg(x)=2(x/2-1/2)+1 fg(x)=x-1+1 fg(x)=x gf(x)=1/2((2x+1)-1) gf(x)=1/2(2x+1-1) gf(x)=x+1/2-1/2 gf(x)=x The functions functions f(x) and g(x) are inverses of each other. This can be demonstarted by f(x)=2x+1 y=2x+1 x=2y+1 x-1=2y (x-1)/2=y Thus, y=1/2(x-1) = g(x) And...
  12. AN630078

    Graphical Transformations and Finding the Equation of a Curve

    Thank you for your reply, sorry I should have used brackets. Question b: y=(4x+17)/(x+2) y=(4x+17)/(x+2)undergoes this first transformation to become y=(4x+17)/(x+2-4) Transform -4 into a fraction, -4(x+2)/(x+2)=(-4x-8)/(x+2) y=(4x+17)/(x+2)(-4x-8)/(x+2) Since the denominators are equal combine...
  13. AN630078

    Graphical Transformations and Finding the Equation of a Curve

    a. y=x^2 undergoes transformation 1 to become y=(x+2)^2 y=x^2+2 undergoes transformation 2 to become y=3(x+2)^2 y=3(x+2)^2 undergoes transformation 3 to become y=3(x+2)^2+4 So would the equation of the resulting curve be y=3(x+2)^2+4? I am very uncertain when it comes to performing...
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