Given that log 3=0.4771 and log 5=0.699, evaluate the following without using logarithm table or calculator . ( a)log135. ( b)log1125​

given that log 3=0.4771 and log 5=0.699, evaluate the following without using logarithm table or calculator . ( a)log135. ( b)log1125​

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  1. Answer is in the attachment below. Please open it up in a new window to see it in full.
    [tex]Evalaute the following without using a calculator: cos(13pi/6)[/tex]

  2. false

    Explanation:

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  3. Answer is in the attachment below. Please open it up in a new window to see it in full.
    [tex]Evalaute the following without using a calculator: cos(13pi/6)[/tex]

  4. [tex]the\ cos\ function\ is\ periodic:\ \ \ the\ period=2 \pi\\\\cos(2 \pi + \alpha )=cos \alpha \\\\------------------------\\\\cos\bigg{(} \frac{\big{13 \pi }}{\big{6}}\bigg{)}=cos\bigg{(} 2 \pi +\frac{\big{ \pi }}{\big{6}}\bigg{)}=cos\bigg{(} \frac{\big{ \pi }}{\big{6}}\bigg{)}= \frac{\big{ \sqrt{3} }}{\big{2}}[/tex]

  5. [tex]\sec x=\dfrac{1}{\cos x}\to \sec60^o=\dfrac{1}{\cos60^o}\\\\\text{Use the table of values of a trigonometric functions}\\\\\cos30^o=\dfrac{\sqrt3}{2}\\\\\sin\dfrac{\pi}{4}=\dfrac{\sqrt2}{2}\\\\\cos60^o=\dfrac{1}{2}\to\sec60^o=\dfrac{1}{\frac{1}{2}}=2\\\\\text{Substitute:}\\\\\cos30^o\sin\dfrac{\pi}{4}+\dfrac{\sec60^o}{3}=\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt2}{2}+\dfrac{2}{3}=\dfrac{\sqrt6}{4}+\dfrac{2}{3}\\\\=\dfrac{3\sqrt6}{3\cdot4}+\dfrac{4\cdot2}{4\cdot3}=\dfrac{3\sqrt6}{12}+\dfrac{8}{12}\\\\=\boxed{\dfrac{8+3\sqrt6}{12}}[/tex]

    [tex]Evaluate the following without using the calculator[tex] \cos(30) \sin( \frac{\pi}{4} ) + \frac{ \se[/tex]

  6. [tex]the\ cos\ function\ is\ periodic:\ \ \ the\ period=2 \pi\\\\cos(2 \pi + \alpha )=cos \alpha \\\\------------------------\\\\cos\bigg{(} \frac{\big{13 \pi }}{\big{6}}\bigg{)}=cos\bigg{(} 2 \pi +\frac{\big{ \pi }}{\big{6}}\bigg{)}=cos\bigg{(} \frac{\big{ \pi }}{\big{6}}\bigg{)}= \frac{\big{ \sqrt{3} }}{\big{2}}[/tex]

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