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ki77a [65]
3 years ago
14

All you need is on the photo

Mathematics
1 answer:
mariarad [96]3 years ago
5 0

Answer:

4^{-3}

Step-by-step explanation:

We\ are\ given\ that,\\4^4*4^{-8}*4\\We\ know\ that,\ for\ any\ real\ number\ n, n^a*n^b=n^{(a+b)}\\Hence,\\First,\ lets\ write\ all\ the\ terms\ in\ an\ exponential\ form.\\4^4*4^{-8}*4^1\\=4^{(4+(-8)+1)}\\=4^{(5-8)}\\=4^{-3}

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Answer:

Step-by-step explanation:

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x = 117/6

x = 19.5 (or 19 1/2) ft <====

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First solve the triangle:
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Prove the following
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Step-by-step explanation:

\large\underline{\sf{Solution-}}

<h2 /><h2><u>Consider</u></h2>

\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \dfrac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \}cos(23π+x)cos(2π+x)

<h2><u>W</u><u>e</u><u> </u><u>K</u><u>n</u><u>o</u><u>w</u><u>,</u></h2>

\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) = sinx

\rm \: {cos \: (2\pi + x) }

\rm \: \cot \bigg( \dfrac{3\pi}{2} - x \bigg) \: = \: tanx

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So, on substituting all these values, we get

\rm \: = \: sinx \: cosx \: (tanx \: + \: cotx)

\rm \: = \: sinx \: cosx \: \bigg(\dfrac{sinx}{cosx} + \dfrac{cosx}{sinx}

\rm \: = \: sinx \: cosx \: \bigg(\dfrac{ {sin}^{2}x + {cos}^{2}x}{cosx \: sinx}

\rm \: = \: 1=1

<h2>Hence,</h2>

\boxed{\tt{ \cos \bigg( \frac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \frac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \} = 1}}

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

<h2>ADDITIONAL INFORMATION :-</h2>

Sign of Trigonometric ratios in Quadrants

  • sin (90°-θ)  =  cos θ
  • cos (90°-θ)  =  sin θ
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  • sec (90°+θ)  =  -csc θ
  • cot (90°+θ)  =  -tan θ
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  • cos (180°-θ)  =  -cos θ
  • tan (180°-θ)  =  -tan θ
  • csc (180°-θ)  =  csc θ
  • sec (180°-θ)  =  -sec θ
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  • sec (270°+θ)  =  cos θ
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