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ycow [4]
3 years ago
12

The lens equation is 1/f=1/p+1/q , where f is the focal length of the lens, p is the distance of the object from the lens, and q

is the distance of the image from the lens. The formula to find q is:
a. q=pf/p-f
b. q=p-f/pf
c. q=pf/p+f
Mathematics
2 answers:
Kisachek [45]3 years ago
6 0

Answer:

q=\frac{pf}{p-f}

Step-by-step explanation:

The lens equation is \frac{1}{f}=\frac{1}{p}+\frac{1}{q}.

To solve this formula for q, we need to multiply each term by fpq.

(fpq)\frac{1}{f}=(fpq)\frac{1}{p}+(fpq)\frac{1}{q}.

We cancel out the common factors to get;

pq=fq+fp.

We group the terms in q on one side of the equation;

pq-fq=fp.

Factor q.

q(p-f)=fp.

Divide both sides by (p-f)

q=\frac{pf}{p-f}

The correct answer is A

vichka [17]3 years ago
5 0
1/f=1/p + 1/q
least comom multiple=pfq
(pq)/(pfq)=(fq)/(pfq)+(pf)/(pfq)
Because all denominators are the same, we can eliminate the denominators.
pq=fq+pf
pq-fq=pf
q(p-f)=pf
q=pf / (p-f)

Answer: a. q=pf / (p-f)

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Step-by-step explanation:

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}5x^3-8x^2+9x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{5x^3}{x}=5x^2\\\mathrm{Quotient}=5x^2\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}5x^2:\:5x^3-15x^2\\\mathrm{Subtract\:}5x^3-15x^2\mathrm{\:from\:}5x^3-8x^2+9x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=7x^2+9x+12\\\mathrm{Therefore}\\=5x^2+\frac{7x^2+9x+12}{x-3}

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}7x^2+9x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{7x^2}{x}=7x\\\mathrm{Quotient}=7x\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}7x:\:7x^2-21x\\\mathrm{Subtract\:}7x^2-21x\mathrm{\:from\:}7x^2+9x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=30x+12\\\mathrm{Therefore}\\=5x^2+7x+\frac{30x+12}{x-3}\\

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}30x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{30x}{x}=30\\\mathrm{Quotient}=30\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}30:\:30x-90\\\mathrm{Subtract\:}30x-90\mathrm{\:from\:}30x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=102\\\mathrm{Therefore}\\=5x^2+7x+30+\frac{102}{x-3}

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Step-by-step explanation:

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