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help_people

  • one year ago

Tabitha wants to hang a painting in a gallery. The painting and frame must have an area of 58 square feet. The painting is 7 feet wide by 8 feet long. Which quadratic equation can be used to determine the thickness of the frame, x? Image of a frame 7 feet wide by 8 feet long, with an of x on the bottom and right sides x2 + 15x − 2 = 0 x2 + 15x + 58 = 0 4x2 + 30x − 2 = 0 4x2 + 30x + 58 = 0

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  1. help_people
    • one year ago
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    @Nishant_Garg

  2. help_people
    • one year ago
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    you here?

  3. anonymous
    • one year ago
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    Hmmm consider the drawing |dw:1442693956662:dw| We are given the combined area(in respective units) \[A_{outer}=58\] and we can find the inner area \[A_{inner}=7 \times 8=56\] then the area of the thickness is \[A_{thickness}=58-56=2\] Now to find an expression for the area of the thickness in terms of x, we can do |dw:1442694478789:dw| This way we have divided out middle section into 4 trapeziums, out of which the left right trapezium are the same and up the down are the same, so they can be just doubled, Now for first pair of trapezium |dw:1442694716457:dw| So we have \[A_{trap1}=\frac{a+b}{2}.h=\frac{8+8+2x}{2}.x=\frac{16+2x}{2}.x=(8+x).x=x^2+8x\] So we have \[2 \times A_{trap1}=2x^2+16x\] Similarly for other pair of trapezium we can do \[2 \times A_{trap2}=(a+b).h=(7+7+2x).x=(14+2x)x=2x^2+14x\] So total Area of thickness or the middle section \[A_{thickness}=2x^2+16x+2x^2+14x=4x^2+30x\] But we know that \[A_{thickness}=2\]\[\therefore 4x^2+30x=2 \implies 4x^2+30x-2=0\]

  4. anonymous
    • one year ago
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    |dw:1442695223716:dw|

  5. help_people
    • one year ago
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    ok

  6. anonymous
    • one year ago
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    or you can simply do \[(8+2x)(7+2x)=58\]\[\implies 4x^2+30x+56=58 \therefore 4x^2+30x-2=0\]

  7. help_people
    • one year ago
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    ty! can oyu help w/ a few mor

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