Question

12 3
(9)
15'5
(10) 규
-7 -3
11' 4​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Step-by-step explanation:

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GIVEN THAT,

b {}^{2} = ac \: \: \: \: \: \: \: \: \: \: \: .....(1)b2=ac.....(1)

Also,

\begin{lgathered}{a}^{x} = {b}^{y} = {c}^{z} \\ \\ let \: \\ {a}^{x} = {b}^{y} = {c}^{z} = p \\ \\ we \: have \\ {a}^{x} = p \\ \\ = > a = {p}^{ \frac{1}{x} } \: \\ \\ similarly \\ b = {p}^{ \frac{1}{y} } \: \: and \: \: c = {p}^{ \frac{1}{z} }\end{lgathered}ax=by=czletax=by=cz=pwehaveax=p=>a=px1similarlyb=py1andc=pz1

Putting value of a, b and c in eq(1)

We get,

\begin{lgathered}{( {p}^{ \frac{1}{y} }) }^{2} = {p}^{ \frac{1}{x} } \times {p}^{ \frac{1}{y} } \\ \\ {p}^{ \frac{2}{y} } = {p}^{ (\frac{1}{x} + \frac{1}{y} )}\end{lgathered}(py1)2=px1×py1py2=p(x1+y1)

On equating the power of p

We get,

\begin{lgathered}\frac{2}{y} = \frac{1}{x} + \frac{1}{z} \\ \\ \frac{2}{y} = \frac{z + x}{xz} \\ \\ \frac{1}{y} = \frac{x + z}{2xz} \\ \\ = > y = \frac{2xz}{x + z} \:\:\:\:\: [Proved] \:\end{lgathered}y2=x1+z1y2=xzz+xy1=2xzx+z=>y=x+z2xz[Proved]