Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

For a 35mm camera, the equation 1/f = 1/di + 1/do represents the focal length (f) of the lens that is needed to produce a clear photograph. ?

Can someone explain to me the step by step on how to solve this? This question genuinely has me stumped.

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  • rotchm
    Lv 7
    1 month ago

    Hints: Unanon yourself and we will further help you. 

    Can you type in the question? It would help of for copy/paste so we can answer you.  

    From 1/f = 1/di + 1/de, thus 1/de = 1/f - 1/di, right?

    Now replace di & f by the given functions. Done!

    Simplify if need be. 

  • 1 month ago

    1/do = 1/f - 1/di

    = (x^2 + x - 2)/(x^2 - 9) - (2x^2 + 10)/(x^2 + 2x - 3)

    = (x^2 + x - 2)(x-1)/[(x+3)(x-3)(x-1)] minus

    (2x^2 + 10)(x - 3)/[(x+3)(x-3)(x-1)]

    = (x^3 - 3x + 2 - 2x^3 + 6x^2 - 10x + 30) / [(x+3)(x-3)(x-1)]

    = (32 - 13x + 6x^2 - x^3)/[(x+3)(x-3)(x-1)]

  • Anonymous
    1 month ago

    Are you that lazy. Some guy answered this last night. 

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