1/TAN(3)[0.928+0.0038(-28.96-4.24)]

asked by guest
on Apr 07, 2025 at 6:56 am



You asked:

Evaluate the expression: 1tan⁡(3)(0.928+0.0038(−28.96−4.24))\frac{1}{\tan\left( 3 \right) \left(0.928 + 0.0038 \left(-28.96 - 4.24\right)\right)}

MathBot Answer:

1tan⁡(3)(0.928+0.0038(−28.96−4.24))=1250010023tan⁡(3)≈−8.74894312011689797145135111683937\frac{1}{\tan\left( 3 \right) \left(0.928 + 0.0038 \left(-28.96 - 4.24\right)\right)} = \frac{12500}{10023 \tan{\left(3 \right)}} \approx -8.74894312011689797145135111683937

1tan⁡(3)(0.928+0.0038(−28.96−4.24))=1tan⁡(3)⋅(9281000+3810000⋅(−(28+96100)−(4+24100)))=1tan⁡(3)⋅(116125+3810000⋅(−(28+96100)−(4+24100)))=1tan⁡(3)⋅(116125+3810000⋅(−(28+2425)−(4+24100)))=1tan⁡(3)⋅(116125+3810000⋅(−72425−(4+24100)))=1tan⁡(3)⋅(116125+3810000⋅(−72425−(4+625)))=1tan⁡(3)⋅(116125+3810000⋅(−72425−10625))=1tan⁡(3)⋅(116125+3810000⋅−1665)=1tan⁡(3)⋅(116125+195000⋅−1665)=1tan⁡(3)⋅(116125−157712500)=1tan⁡(3)⋅1002312500=11002312500⋅tan⁡(3)=1250010023⋅1tan⁡(3)\begin{aligned}\frac{1}{\tan\left( 3 \right) \left(0.928 + 0.0038 \left(-28.96 - 4.24\right)\right)}&=\frac{1}{\tan{\left(3 \right)} \cdot \left(\frac{928}{1000} + \frac{38}{10000} \cdot \left(- \left(28 + \frac{96}{100}\right) - \left(4 + \frac{24}{100}\right)\right)\right)}\\&=\frac{1}{\tan{\left(3 \right)} \cdot \left(\frac{116}{125} + \frac{38}{10000} \cdot \left(- \left(28 + \frac{96}{100}\right) - \left(4 + \frac{24}{100}\right)\right)\right)}\\&=\frac{1}{\tan{\left(3 \right)} \cdot \left(\frac{116}{125} + \frac{38}{10000} \cdot \left(- \left(28 + \frac{24}{25}\right) - \left(4 + \frac{24}{100}\right)\right)\right)}\\&=\frac{1}{\tan{\left(3 \right)} \cdot \left(\frac{116}{125} + \frac{38}{10000} \cdot \left(- \frac{724}{25} - \left(4 + \frac{24}{100}\right)\right)\right)}\\&=\frac{1}{\tan{\left(3 \right)} \cdot \left(\frac{116}{125} + \frac{38}{10000} \cdot \left(- \frac{724}{25} - \left(4 + \frac{6}{25}\right)\right)\right)}\\&=\frac{1}{\tan{\left(3 \right)} \cdot \left(\frac{116}{125} + \frac{38}{10000} \cdot \left(- \frac{724}{25} - \frac{106}{25}\right)\right)}\\&=\frac{1}{\tan{\left(3 \right)} \cdot \left(\frac{116}{125} + \frac{38}{10000} \cdot - \frac{166}{5}\right)}\\&=\frac{1}{\tan{\left(3 \right)} \cdot \left(\frac{116}{125} + \frac{19}{5000} \cdot - \frac{166}{5}\right)}\\&=\frac{1}{\tan{\left(3 \right)} \cdot \left(\frac{116}{125} - \frac{1577}{12500}\right)}\\&=\frac{1}{\tan{\left(3 \right)} \cdot \frac{10023}{12500}}\\&=\frac{1}{\frac{10023}{12500} \cdot \tan{\left(3 \right)}}\\&=\frac{12500}{10023} \cdot \frac{1}{\tan{\left(3 \right)}}\end{aligned}

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