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Question-223242




Question Number 223242 by mnjuly1970 last updated on 19/Jul/25
Answered by Tawa11 last updated on 19/Jul/25
Sir, did you hear from     Mathdave???  Or you know where he is?
$$\mathrm{Sir},\:\mathrm{did}\:\mathrm{you}\:\mathrm{hear}\:\mathrm{from}\:\:\:\:\:\mathrm{Mathdave}??? \\ $$$$\mathrm{Or}\:\mathrm{you}\:\mathrm{know}\:\mathrm{where}\:\mathrm{he}\:\mathrm{is}? \\ $$
Commented by mnjuly1970 last updated on 19/Jul/25
  hello sir . no , i didn′t hear it .
$$\:\:{hello}\:{sir}\:.\:{no}\:,\:{i}\:{didn}'{t}\:{hear}\:{it}\:. \\ $$
Answered by mr W last updated on 19/Jul/25
Commented by mr W last updated on 19/Jul/25
(x/(LM))=(b/(LN))   ...(i)  (x/(LN))=(a/(LM))   ...(ii)  (i)×(ii):  x^2 =ab
$$\frac{{x}}{{LM}}=\frac{{b}}{{LN}}\:\:\:…\left({i}\right) \\ $$$$\frac{{x}}{{LN}}=\frac{{a}}{{LM}}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$${x}^{\mathrm{2}} ={ab} \\ $$
Commented by mnjuly1970 last updated on 19/Jul/25
 ⋛
$$\:\cancel{\underline{\underbrace{\lesseqgtr}}} \\ $$
Answered by mr W last updated on 19/Jul/25
Commented by mr W last updated on 19/Jul/25
(x/(LM))=(b/(LN))    ...(i)  (x/(LN))=(a/(LM))    ...(ii)  (i)×(ii):  x^2 =ab
$$\frac{{x}}{{LM}}=\frac{{b}}{{LN}}\:\:\:\:…\left({i}\right) \\ $$$$\frac{{x}}{{LN}}=\frac{{a}}{{LM}}\:\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$${x}^{\mathrm{2}} ={ab} \\ $$
Commented by mnjuly1970 last updated on 19/Jul/25
  grateful  sir  W
$$\:\:{grateful}\:\:{sir}\:\:{W} \\ $$
Answered by mr W last updated on 19/Jul/25
Commented by mr W last updated on 19/Jul/25
(a/p)=(h/r)=(b/q) ⇒(p/q)=(a/b)  ((x−a)/p)=((b−x)/q)  ((x−a)/(b−x))=(p/q)=(a/b)  ⇒2ab=x(a+b)  ⇒(2/x)=(1/a)+(1/b) ✓
$$\frac{{a}}{{p}}=\frac{{h}}{{r}}=\frac{{b}}{{q}}\:\Rightarrow\frac{{p}}{{q}}=\frac{{a}}{{b}} \\ $$$$\frac{{x}−{a}}{{p}}=\frac{{b}−{x}}{{q}} \\ $$$$\frac{{x}−{a}}{{b}−{x}}=\frac{{p}}{{q}}=\frac{{a}}{{b}} \\ $$$$\Rightarrow\mathrm{2}{ab}={x}\left({a}+{b}\right) \\ $$$$\Rightarrow\frac{\mathrm{2}}{{x}}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\:\checkmark \\ $$

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