1/TAN(3)[0.928+0.0038(-16-12.49)]

asked by guest
on Apr 07, 2025 at 6:55 am



You asked:

Evaluate the expression: 1tan⁡(3)(0.928+0.0038(−16−12.49))\frac{1}{\tan\left( 3 \right) \left(0.928 + 0.0038 \left(-16 - 12.49\right)\right)}

MathBot Answer:

1tan⁡(3)(0.928+0.0038(−16−12.49))=500000409869tan⁡(3)≈−8.5579203982669261025212340766519\frac{1}{\tan\left( 3 \right) \left(0.928 + 0.0038 \left(-16 - 12.49\right)\right)} = \frac{500000}{409869 \tan{\left(3 \right)}} \approx -8.5579203982669261025212340766519

1tan⁡(3)(0.928+0.0038(−16−12.49))=1tan⁡(3)⋅(9281000+3810000⋅(−16−(12+49100)))=1tan⁡(3)⋅(116125+3810000⋅(−16−(12+49100)))=1tan⁡(3)⋅(116125+3810000⋅(−16−1249100))=1tan⁡(3)⋅(116125+3810000⋅−2849100)=1tan⁡(3)⋅(116125+195000⋅−2849100)=1tan⁡(3)⋅(116125−54131500000)=1tan⁡(3)⋅409869500000=1409869500000⋅tan⁡(3)=500000409869⋅1tan⁡(3)\begin{aligned}\frac{1}{\tan\left( 3 \right) \left(0.928 + 0.0038 \left(-16 - 12.49\right)\right)}&=\frac{1}{\tan{\left(3 \right)} \cdot \left(\frac{928}{1000} + \frac{38}{10000} \cdot \left(-16 - \left(12 + \frac{49}{100}\right)\right)\right)}\\&=\frac{1}{\tan{\left(3 \right)} \cdot \left(\frac{116}{125} + \frac{38}{10000} \cdot \left(-16 - \left(12 + \frac{49}{100}\right)\right)\right)}\\&=\frac{1}{\tan{\left(3 \right)} \cdot \left(\frac{116}{125} + \frac{38}{10000} \cdot \left(-16 - \frac{1249}{100}\right)\right)}\\&=\frac{1}{\tan{\left(3 \right)} \cdot \left(\frac{116}{125} + \frac{38}{10000} \cdot - \frac{2849}{100}\right)}\\&=\frac{1}{\tan{\left(3 \right)} \cdot \left(\frac{116}{125} + \frac{19}{5000} \cdot - \frac{2849}{100}\right)}\\&=\frac{1}{\tan{\left(3 \right)} \cdot \left(\frac{116}{125} - \frac{54131}{500000}\right)}\\&=\frac{1}{\tan{\left(3 \right)} \cdot \frac{409869}{500000}}\\&=\frac{1}{\frac{409869}{500000} \cdot \tan{\left(3 \right)}}\\&=\frac{500000}{409869} \cdot \frac{1}{\tan{\left(3 \right)}}\end{aligned}

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