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loris [4]
1 year ago
10

Hello! Please help me with this. I'm very desperate lol. Its 16. Ill give brainliest

Mathematics
1 answer:
Paladinen [302]1 year ago
7 0

Answer:

You showed the right work and got the right answer!

x = 6

Step-by-step explanation:

Since B is in the middle of A and C, AB + BC = AC

So,

2x - 11 + 8 = x + 3

x - 3 = 3

x = 6

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Simplify: <br>{(12)^1 + (13)^-1}/[(1/5)^-2 × {(1/5)^-1 + (1/8)^-1}^-1]​
CaHeK987 [17]

Step-by-step explanation:

\underline{\underline{\sf{➤\:\:Solution}}}

\sf \dashrightarrow \:  \dfrac{ \left(\left(12 \right)^{ - 1}  + \left(13 \right)^{ - 1}  \right) }{\left( \dfrac{1}{5}\right) ^{ - 2}  \times\left( \left( \dfrac{1}{5}  \right) ^{ - 1}  +\left( \dfrac{1}{8}  \right) ^{ - 1}  \right) ^{ - 1}}

\sf \dashrightarrow \:  \dfrac{ \left(\dfrac{1}{12}  + \dfrac{1}{13} \right) }{\left( \dfrac{5}{1}\right) ^{ 2}  \times\left( \dfrac{5}{1}  + \dfrac{8}{1}   \right) ^{ - 1}}

  • LCM of 12 and 13 is 156

\sf \dashrightarrow \:  \dfrac{ \left(\dfrac{1 \times 13 = 13}{12 \times 13 = 156}  + \dfrac{1 \times 12 = 12}{13 \times 12 = 156} \right) }{\ \dfrac{25}{1} \times\left( \dfrac{5 + 8}{1}    \right) ^{ - 1}}

\sf \dashrightarrow \:  \dfrac{ \left(\dfrac{13}{156}  + \dfrac{12}{156} \right) }{\ \dfrac{25}{1} \times\left( \dfrac{13}{1}    \right) ^{ - 1}}

\sf \dashrightarrow \:  \dfrac{ \left(\dfrac{13 + 12}{156}  \right) }{\ \dfrac{25}{1} \times\dfrac{1}{13} }

\sf \dashrightarrow \:    \dfrac{25}{156} \div    \dfrac{25}{13}

\sf \dashrightarrow \:    \dfrac{ \cancel{25}}{156}  \times    \dfrac{13}{ \cancel{25} }

\sf \dashrightarrow \:     \dfrac{13}{156}

\sf \dashrightarrow \:     \dfrac{1}{12}

\sf \dashrightarrow \:    Answer =   \underline{\boxed{ \sf{ \dfrac{1}{12} }}}

━━━━━━━━━━━━━━━━━━━━━━━━

\underline{\underline{\sf{★\:\:Laws\:of\: Exponents :}}}

\sf \: 1^{st} \: Law = \bigg( \dfrac{m}{n} \bigg)^{a} \times \bigg( \dfrac{m}{n} \bigg)^{b} = \bigg( \dfrac{m}{n} \bigg)^{a + b}

\sf 2^{nd} \: Law =

\sf Case : (i) \: if \: a > b \: then, \bigg( \dfrac{m}{n}\bigg) ^{a} \div \bigg( \dfrac{m}{n}\bigg)^{b} = \bigg( \dfrac{m}{n}\bigg)^{a - b}

\sf Case : (ii) \: if \: a < b \: then, \bigg( \dfrac{m}{n}\bigg) ^{a} \div \bigg( \dfrac{m}{n}\bigg)^{b} = \dfrac{1}{\bigg( \dfrac{m}{n}\bigg)^{b - a}}

\sf \: 3^{rd} \: Law = \bigg\{ \bigg( \dfrac{m}{n} \bigg)^{a} \bigg\}^{b} = \bigg( \dfrac{m}{n} \bigg)^{a \times b} =\bigg( \dfrac{m}{n} \bigg)^{ab}

\sf \: 4^{th} \: Law = \bigg( \dfrac{m}{n} \bigg)^{ - 1} = \bigg( \dfrac{n}{m} \bigg) =\dfrac{n}{m}

\sf \: 5^{th} \: Law = \bigg( \dfrac{m}{n} \bigg)^{0} = 1

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