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anonymous

  • 5 years ago

A 10-ft diagonal brace on a bridge connects a support of the center of the bridge to a side support on the bridge. The horizontal distance that it spans is 2 ft longer that the height that it reaches on the side of the bridge. Find the horizontal and vertical distances spanned by this brace.

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  1. anonymous
    • 5 years ago
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    This is what i have so far (x + 2)^2 + x^2 = 100 X^2 +4 + 2 (x) (2) +x^2 = 100 Using, (a + b)^2 = a^2 + b^2 +2ab 2x^2 +4x -96 = 0 So x^2 + 2x -48 = 0 Its 8 feet and 6 feet. X is 6 ft. Using this formula, Ax^2 +bx +c = 0 x = -b ± √(b^2- 4ac)/2a 8=-b±√(b^2-4ac)/2a

  2. anonymous
    • 5 years ago
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    from here I have no clue.

  3. anonymous
    • 5 years ago
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    or am i wrong.

  4. anonymous
    • 5 years ago
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    Lol Look at the pythagorean thereom Ill send a pic with it Solve for y [4y^2 + y^2] = (5^1/2)*y = 10 y = 10/(5)^1/2 = 4.5 ft (vertical) and 9.0ft (horizontal)

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  5. anonymous
    • 5 years ago
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    I am sorry that is a mistake...I made the horizontal distance 2 times the vertical new solution in a minute

  6. anonymous
    • 5 years ago
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    ok

  7. anonymous
    • 5 years ago
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    u there.

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