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.
- rotchmLv 71 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.
- az_lenderLv 71 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)]
- Anonymous1 month ago
Are you that lazy. Some guy answered this last night.