Question

Solve this:-
$\frac{11 {}^{z - 2} - 99 \times 11 {}^{z - 1} }{143 \times 11 {}^{z - 1} + {11}^{z + 1} \times 9 }$
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Answer 1

Step-by-step explanation:

Given Question:-

Solve [11^(z-1)-99×11^(z-1)]/[143×11^(z-1)+11^(z+1)×9]

Solution:-

See above attachment for understanding

Given that

[11^(z-1)-99×11^(z-1)]/[143×11^(z-1)+11^(z+1)×9]

We know that

a^(m-n)=a^m/a^n

a^(m+n)=a^m×a^n

[(11^z)/(11^2)-99×(11^z)/(11)]/[143×(11^z)/(11)+11^z×11×9]

[(11^z/121)-9×11^z]/[(13×11^z)+(11^z×99)]

11^z[(1/121)-9]/11^z(13+99)

On cancelling 11^z then

(1/121-9)/(112)

[1-(121×9)/121]/112

[(1-1089)/121]/112

(-1088)/(121×112)

(-68)/(121×7)

-68/847

Answer:-

The answer for the given problem is -68/847

Used formulae:-

• a^(m-n)=a^m/a^n
• a^(m+n)=a^m×a^n

Answer 2

$\large\underline{\bold{Given \:Question - }}$

$\tt \: Solve \: : \: \dfrac{11 {}^{z - 2} - 99 \times 11 {}^{z - 1} }{143 \times 11 {}^{z - 1} + {11}^{z + 1} \times 9 }$

$\large\underline{\bold{Solution-}}$

Basic Concept Used :-

We have to first reduce the terms to simplest form and then take out common and use the law of exponents to simply this.

Laws of exponents :-

$\begin{gathered}(1)\:{\underline{\boxed{\bf{{a^m\times{a^n}\:=\:a^{m\:+\:n}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(2)\:{\underline{\boxed{\bf{{\dfrac{a^m}{a^n}\:=\:a^{m\:-\:n}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(3)\:{\underline{\boxed{\bf{{\dfrac{1}{x^n}\:=\:x^{-n}\:}}}}} \\ \end{gathered}$

Let's do it now!!

The given statement is

$\rm :\longmapsto\: \dfrac{11 {}^{z - 2} - 99 \times 11 {}^{z - 1} }{143 \times 11 {}^{z - 1} + {11}^{z + 1} \times 9 }$

$\rm :\longmapsto\: \dfrac{11 {}^{z - 2} - 11 \times 9 \times 11 {}^{z - 1} }{ 13 \times 11 \times 11 {}^{z - 1} + {11}^{z + 1} \times {3}^{2} }$

$\rm :\longmapsto\: \dfrac{11 {}^{z - 2} - 9 \times 11 {}^{z - 1 + 1} }{ 13\times 11 {}^{z - 1 + 1} + {11}^{z + 1} \times {3}^{2} }$

$\rm :\longmapsto\: \dfrac{11 {}^{z - 2} - 9 \times 11 {}^{z} }{ 13 \times 11 {}^{z} + {11}^{z + 1} \times {3}^{2} }$

$\rm :\longmapsto\:\dfrac{ {11}^{z} \times ( { {11}^{ - 2} - 9)}}{ {11}^{z} \times ( 13 + 11 \times 9)}$

$\rm :\longmapsto\:\dfrac{\dfrac{1}{121} - 9 }{ (13 + 99)}$

$\rm :\longmapsto\:\dfrac{1 - 1089}{121 \times 3 \times (4 + 33)}$

$\rm :\longmapsto\:\dfrac{ - 1088}{121 \times 112}$

$\rm :\longmapsto\:\dfrac{ - 1088}{13552}$

= -68/847