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YanaSidlinskiy

  • one year ago

The floor of a shed has an area of 99 square feet. The floor is in the shape of a rectangle whose length is 7 feet less than twice the width. Find the length and the width of the floor of the shed. Use the formula, area=length*width

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  1. YanaSidlinskiy
    • one year ago
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    length is 7 feet less than twice the width = 7-2x? Something like that...

  2. mathstudent55
    • one year ago
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    width = w twice the width: 2w 7 ft less than twice the width: 2w - 7

  3. mathstudent55
    • one year ago
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    A = LW = w(2w - 7) = 99

  4. YanaSidlinskiy
    • one year ago
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    WAIT! I want to solve it lol. I need to understand it on my own on how to do it haha. Wait, why would it matter if I put it the way you showed me?

  5. mathstudent55
    • one year ago
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    There is a difference in the way you write it. Let me explain. Let's say Mary is 16 years old and Jane is 18 years old. You can say Mary's age is 2 years less than Jane's age. If that means 18 - 2, then Mary is 16 years old which is correct. If mistakenly you think that means 2 - 18 = -16, then Mary is -16 year old which is meaningless.

  6. mathstudent55
    • one year ago
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    "7 less than 2w" is 2w - 7 "7 less" means subtract 7, take away 7, from 2w 7 - 2w is "2w less than 7"

  7. YanaSidlinskiy
    • one year ago
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    Gotcha! Ok, makes sense. Lemme solve this real quick. Sorry if it'll take a while..

  8. mathstudent55
    • one year ago
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    No problem. Take your time.

  9. YanaSidlinskiy
    • one year ago
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    I got 22.

  10. mathstudent55
    • one year ago
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    w(2w - 7) = 99 2w^2 - 7w - 99 = 0

  11. YanaSidlinskiy
    • one year ago
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    That's what I had in the first place..

  12. mathstudent55
    • one year ago
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    \(\large x = \dfrac{-b \pm \sqrt {b^2 - 4ac}}{2a} \) \(\large w = \dfrac{7 \pm \sqrt {(-7)^2 - 4(2)(-99)}}{2(2)} \) \(\large w = \dfrac{7 \pm \sqrt {49 + 792}}{4} \) \(\large w = \dfrac{7 \pm \sqrt {841}}{4} \) \(\large w = \dfrac{7 \pm 29}{4} \) \(\large w = \dfrac{36}{4} \) or \(w = \dfrac{-22}{4}\) \(\large w = 9\) or \(\large w = -5.5\)

  13. mathstudent55
    • one year ago
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    Since a width cannot be negative, we discard w = -5.5. The width is 9 ft. The length is 2w - 7 = 2(9) - 7 = 18 - 7 = 11 ft

  14. YanaSidlinskiy
    • one year ago
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    It's the minor stuff that kills me lol.

  15. mathstudent55
    • one year ago
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    You need to work carefully. I went through the same, making stupid mistakes. Do your work and check it.

  16. mathstudent55
    • one year ago
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    When you find an answer, check the answer. We get W = 9 ft and L = 11 ft Let's see if it makes sense. The area has to be 99 ft^2 A = LW = 11 ft * 9 ft = 99 ft^2 The length has to be 7 less than twice the width. L = 2w - 7 = 2(9) - 7 = 18 - 7 = 11 ft Yes, L = 11 ft and W = 9 ft make sense.

  17. YanaSidlinskiy
    • one year ago
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    Gotcha! Ok, makes sense!

  18. YanaSidlinskiy
    • one year ago
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    Thanks for the detailed explanations. I appreciate it!

  19. mathstudent55
    • one year ago
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    You are very welcome.

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