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anonymous

  • one year ago

Part A:The area of a square is (4x2 − 12x + 9) square units. Determine the length of each side of the square by factoring the area expression completely. Show your work. (5 points) Part B: The area of a rectangle is (16x2 − 9y2) square units. Determine the dimensions of the rectangle by factoring the area expression completely. Show your work. (5 points)

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  1. anonymous
    • one year ago
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    @Mackenzie_Willa

  2. anonymous
    • one year ago
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    @welshfella

  3. anonymous
    • one year ago
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    I have the answer just not the steps! Part A:(2x -3)^2 Part B:(4x+3y)(4x−3y)

  4. anonymous
    • one year ago
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    @Nnesha

  5. amilapsn
    • one year ago
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    Do you know these? \[a^2-b^2=(a-b)(a+b) \\ (a+b)^2=a^2+2ab+b^2\]

  6. anonymous
    • one year ago
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    Because the first shape is a square, the length and the width must be the same. So the area must be a perfect square. We need to find an expression that , when multiplied by itself, will give \(4x^2-12x+9\)Start by taking the square roots of the first and last coefficients. Can you do that?

  7. anonymous
    • one year ago
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    Can you help me work it ut step by step plz

  8. anonymous
    • one year ago
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    *out

  9. anonymous
    • one year ago
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    What is the square root of the first term, \(4x^2\)?

  10. anonymous
    • one year ago
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    2x^2?

  11. anonymous
    • one year ago
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    Not quite. The square of 4 is 2, so you got that part right, but what is the square root of x^2? What can you multiply by itself to give x^2?

  12. anonymous
    • one year ago
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    *square root of 4

  13. anonymous
    • one year ago
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    I am confused

  14. anonymous
    • one year ago
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    Do you understand that\[x \times x = x^2\]

  15. anonymous
    • one year ago
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    yes

  16. anonymous
    • one year ago
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    OK. So the square root of \(x^2\) is \(x\). OK with that?

  17. anonymous
    • one year ago
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    oh okay.

  18. anonymous
    • one year ago
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    All right. So the square root of \(4x^2\) is \(2x\). In other words\[2x \times 2x = 4x^2\]Make sense now?

  19. anonymous
    • one year ago
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    Yes

  20. anonymous
    • one year ago
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    OK. So now you need the square root of the last term, +9. What is it?

  21. anonymous
    • one year ago
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    3

  22. anonymous
    • one year ago
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    Great. With these two square roots, you have determined that the two factors must be\[4x^2-12x+9 = \left( 2x \text{ ? }3 \right)\left( 2x \text{ ? }3 \right)\]All that is left is to choose the correct sign, plus or minus. For that, look at the second term in the given trinomial. What sign does it have?

  23. anonymous
    • one year ago
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    Minus.

  24. anonymous
    • one year ago
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    Perfect. Then that is the sign you use in the factors. Therefore,\[4x^2-12x+9 = \left( 2x-3 \right)\left( 2x-3 \right)\]and you have your answer.

  25. anonymous
    • one year ago
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    Keep in mind that this technique only works for factoring a perfect square trinomial.

  26. anonymous
    • one year ago
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    Are these the answers. I have been working them out too: Part A) 4x2 − 12x + 9 4x^2 -6x -6x + 9 2x( 2x - 3) -3(2x - 3) (2x - 3)(2x-3) length of each side is 2x-3 Part B) 16x^2 − 9y^2 (4x)^2 - (3y)^2 (4x - 3y) (4x+3y) dimensions are 4x-3y, 4x+3y.

  27. anonymous
    • one year ago
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    Yes, you are correct. Very well done!

  28. anonymous
    • one year ago
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    Thank you!

  29. anonymous
    • one year ago
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    You're welcome.

  30. anonymous
    • one year ago
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    @ospreytriple can you help with 1 more?

  31. anonymous
    • one year ago
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    An expression is shown below: f(x) = –16x2 + 60x + 16 Part A: What are the x-intercepts of the graph of the f(x)? Show your work. (2 points) Part B: Is the vertex of the graph of f(x) going to be a maximum or minimum? What are the coordinates of the vertex? Justify your answers and show your work. (3 points) Part C: What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part A and Part B to draw the graph. (5 points)

  32. anonymous
    • one year ago
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    Part A: To find the x-intercepts, you must solve the equation \(-16x^2+60x+16=0\). Which method would you use; factoring by completing the square, or the quadratic formula?

  33. anonymous
    • one year ago
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    quadratic formula?

  34. anonymous
    • one year ago
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    OK. Go ahead. What do you get?

  35. anonymous
    • one year ago
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    Can you refresh me on the formula.

  36. anonymous
    • one year ago
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    For the general quadratic, \(ax^2 + bx + c=0\), the x-intercepts are found using\[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

  37. anonymous
    • one year ago
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    can you help me through this one

  38. anonymous
    • one year ago
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    I am really confused.

  39. anonymous
    • one year ago
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    Have you used the quadratic formula before?

  40. anonymous
    • one year ago
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    Just learning it.

  41. anonymous
    • one year ago
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    OK. I'll help you set it up.

  42. anonymous
    • one year ago
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    Okay thank you! I do have 1 request can you say like part a and like part b and part c because i will get confused and don't know what to say.

  43. anonymous
    • one year ago
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    This is for Part A. If you look at the given trinomial and compare it with the general quadratic, you'll see that \(a=-16\), \(b=60\) and \(c=16\). Putting these values into the quadratic formula, you have\[x=\frac{ -b \pm \sqrt{b^2-4ac} }{2a } = \frac{ -60 \pm \sqrt{60^2-4\left( -16 \right)\left( 16 \right)} }{ 2\left( -16 \right) }\]Can you complete these calculations?

  44. anonymous
    • one year ago
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    is it -5?

  45. anonymous
    • one year ago
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    One second. I'm working it out...

  46. anonymous
    • one year ago
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    okay ;]

  47. anonymous
    • one year ago
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    Wait is it... x = 4 and x = -1/4.

  48. anonymous
    • one year ago
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    Yayyyy! You got it. Those are the x-intercepts, \(x=4\) and \(x=-\frac{1}{4}\)

  49. anonymous
    • one year ago
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    Yay :V. can you help with part B and C too?

  50. anonymous
    • one year ago
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    Part B: Vertex a maximum or minimum? What it means is, is the parabola opening upward or downward? If the parabola opens upward, then the vertex is a minimum. If the parabola opens downward, then the vertex is a maximum.|dw:1440091745676:dw| Do you know how to tell which direction the parabola opens?

  51. anonymous
    • one year ago
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    These are my answers so far are they correct? Part A.) f(x) = –16x^2 + 60x + 16 -4(x-4)(x+1/4) x-4=0 x+1/4=0 x=4 and x=-1/4 Part B.) vertex (15/8, 289/4) it is a maximum because the coeffficient of x^2 is -16 which is negative Part C.) You can graph the x intercepts and the vertex. That will give you the overall shape of the function.

  52. anonymous
    • one year ago
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    OK. That saves me a lot of work. Hope I'm not wasting your time. I see in Part A you factored instead of using the quadratic formula. Parts B & C are correct. Good job>

  53. anonymous
    • one year ago
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    Part c is correct?

  54. anonymous
    • one year ago
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    Yes.

  55. anonymous
    • one year ago
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    Oh. I wrote that down for notes rofl

  56. anonymous
    • one year ago
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    Thank you!

  57. anonymous
    • one year ago
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    You're welcome

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