Please solve the attached question
I will mark the correct answer and explanation as brainliest
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Please solve the attached question
I will mark the correct answer and explanation as brainliest
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Answer:
AnswEr :
⠀⠀⠀⌬ Let the Unit's place digit = M
⠀⠀⠀⌬ And the ten's place digit = N
⠀⠀⠀⌬ Then, Original Number = 10(M + N)
⠀⠀⠀⌬ Also, Interchaged Number = 10(N + M)
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• F i r s t⠀C o n d i t i o n :
Sum of the two digits (M + N) number is 9.
\begin{gathered}\twoheadrightarrow\sf M + N = 9 \qquad\quad\qquad\quad\Bigg\lgroup\sf eq^{n}\;(1)\Bigg\rgroup\\\\\end{gathered}
↠M+N=9
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• S e c o n d⠀C o n d i t i o n :
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After Interchanging the digits, the new number is greater than the original number by 27.
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\begin{gathered}\longrightarrow\sf (10N + M) - (10M + N) = 27\\\\\\\end{gathered}
⟶(10N+M)−(10M+N)=27
\begin{gathered}\longrightarrow\sf 9N - 9M = 27\\\\\\\end{gathered}
⟶9N−9M=27
\begin{gathered}\longrightarrow\sf N - M = 3\\\\\\\end{gathered}
⟶N−M=3
\begin{gathered}\longrightarrow\sf N = 3 + M \qquad\quad\qquad\quad\Bigg\lgroup\sf eq^{n}\;(2)\Bigg\rgroup\\\\\end{gathered}
⟶N=3+M
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\begin{gathered}\underline{\bigstar\:\sf{Substitue \: the \: value \: of \: N \: from \: eq^n \: (2) \: to \: eq^n \: (1) : }} \\ \\ \\ \end{gathered}
★SubstituethevalueofNfromeq
n
(2)toeq
n
(1):
\begin{gathered}\longrightarrow\sf M + N = 9\\\\\\\end{gathered}
⟶M+N=9
\begin{gathered}\longrightarrow\sf M + 3 + M = 9\\\\\\\end{gathered}
⟶M+3+M=9
\begin{gathered}\longrightarrow\sf 2M = 9 - 3 \\\\\\\end{gathered}
⟶2M=9−3
\begin{gathered}\longrightarrow\sf 2M = 6\\\\\\\end{gathered}
⟶2M=6
\begin{gathered}\longrightarrow\sf M = \cancel\dfrac{6}{2}\\\\\\\end{gathered}
⟶M=
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6
\begin{gathered}\longrightarrow\sf M = 3\\\\\end{gathered}
⟶M=3
\begin{gathered}\underline{\bigstar\:\sf{Substituting\;value\;of\;M\;in\;\;eq^n\;(1)\;: }} \\ \\ \\ \end{gathered}
★SubstitutingvalueofMineq
n
(1):
\begin{gathered}\longrightarrow\sf M + N = 9\\\\\\\end{gathered}
⟶M+N=9
\begin{gathered}\longrightarrow\sf 3 + N = 9\\\\\\\end{gathered}
⟶3+N=9
\begin{gathered}\longrightarrow\sf N = 9 - 3\\\\\\\end{gathered}
⟶N=9−3
\begin{gathered}\longrightarrow\sf N = 6\\\\\end{gathered}
⟶N=6
\begin{gathered}\underline{\bigstar\:\textsf{Now,\;Original\; Number\; :}}\\\\\end{gathered}
★Now,OriginalNumber:
\begin{gathered}\twoheadrightarrow\sf Original\; Number = 10(M + N)\\\\\\\end{gathered}
↠OriginalNumber=10(M+N)
\begin{gathered}\twoheadrightarrow\sf Original\; Number = 10(3) + 6\\\\\\\end{gathered}
↠OriginalNumber=10(3)+6
\begin{gathered}\twoheadrightarrow\sf Original\; Number = 30 + 6\\\\\\\end{gathered}
↠OriginalNumber=30+6
\begin{gathered}\twoheadrightarrow\underline{\boxed{\pmb{\sf{ Original\; Number =36}}}}\\\\\end{gathered}
↠
OriginalNumber=36
OriginalNumber=36
\;\;\;\;\;\qquad\therefore{\underline{\textsf{Hence, the Original number is \textbf{36}.}}}∴
Hence, the Original number is 36.
Verified answer
[tex]\huge{⚝}[/tex][tex]\huge{\underline{\mathcal{Answer}}}\\[/tex]
[tex]\huge{☞}[/tex][tex]\large{\sf k = 0}[/tex]
[tex]\\\\\huge{⚝}[/tex][tex]\\\\\huge{\underline{\mathcal{Given:}}}[/tex]
[tex]f(x) = \begin{cases} \dfrac{1-cos x}{x} & \text{, } x \neq 0 \\ k & \text{, x = 0}\end{cases}[/tex]
is continuous at x = 0.
[tex]\\\\[/tex]
[tex]\huge{⚝}[/tex][tex]\huge{\underline{\mathcal{To Find:}}}\\[/tex]
[tex]\huge{→}[/tex] Value of k
[tex]\\\\[/tex]
[tex]\huge{⚝}[/tex][tex]\huge{\underline{\mathcal{Steps}}}\\[/tex]
Here, f(0) = k
For f to be continuous f(0)=[tex]\lim_{x \to 0}[/tex]f(x).
[tex]\\[/tex]
[tex]\large{⇒}[/tex]k = [tex]\lim_{x \to 0}[/tex]f(x)[tex]\\[/tex]
[tex]\large{⇒}[/tex]k = [tex]\lim_{x \to 0} \dfrac{1- \cos x}{x}\\[/tex]
Differentiating numerator and denominator,
[tex]\large{⇒}[/tex]k = [tex]\lim_{x \to 0} \dfrac{0-(-\sin x)}{1}\\[/tex]
[tex]\large{⇒}[/tex]k = [tex]\lim_{x \to 0} \sin x\\[/tex]
[tex]\large{⇒}[/tex]k = [tex]\lim_{x \to 0} \sin 0 \\[/tex]
[tex]\large{⇒}[/tex]k = 0
[tex]\\\\\fbox{\fbox{\fbox{\fbox{\huge{\underline{\underline{\sf{\green{Hope\:it\:helps}}}}}}}}}[/tex]