Math, asked by Anonymous, 2 months ago

Please solve this question.
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arnab16062008: ok
arnab16062008: itzcute bro
Anonymous: Tume thanks bolne ki jrurt nhi h le♥️
Anonymous: are harsh bolo bro
arnab16062008: I have 13 thx to you
Anonymous: bro 13♥️=50 thanks . hehe
Anonymous: hehe
Anonymous: (◕ᴗ◕✿)
mandeep669: hi
Anonymous: hi

Answers

Answered by Itzraisingstar
80

\huge\fcolorbox{black}{lime}{AnsweR:}

\large\boxed{\mathfrak{2^9*(\frac{1}{3})^1^7 }}

\bold{ExplantioN:}

\large\boxed{\mathfrak{\frac{(-2)^2^4*(-3)^1^7*(8)^-^9}{(12)^1^6*(27)^6} }}

\large\boxed{\mathfrak{\frac{(-2^2^4)*(-3)^1^7*(8)^-^9}{(12)^1^6*(3^3)^6} }}

\large\boxed{\mathfrak{\frac{(-2^2^4)*(-3)^1^7*(8)^-^9}{(12)^1^6*(3)^1^8} }}\\\\\\\large\boxed{\mathfrak{\frac{(-1*2)^2^4*(-1*3)^1^7*2^-^2^7}{2^1^6*2^1^6*3^1^6*3^1^8} }}

\large\boxed{\mathfrak{\frac{-1^2^4^+^1^7*2^2^4^+^1^7*3^1^7}{2^3^2*3^3^4} }}

\large\boxed{\mathfrak{-1^4^1*2^9*3^-^1^7}}

\large\boxed{\mathfrak{2^9*(\frac{1}{3})^1^7 }}      .

❌No Spam allowed❌​

But sorry I'll spam and I'll go against you for me everything is allowed :)

\huge\boxed{\mathfrak{30thanks+follow=inbox}}

Answered by Anonymous
64

\huge \mathbb{ \red {★᭄ꦿ᭄S} \pink{ᴏ}\purple{ʟᴜ} \blue {ᴛ} \orange{ɪ} \green{ᴏɴ★᭄ꦿ᭄}}

\huge{ \mathfrak{ \red{find \: } \purple{attach} \pink{ment}}}

\boxed{\huge \star{{\mathfrak{\red{☠Hãrshü \: : hêrë☠}\star}}}}

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