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Sedaia [141]
2 years ago
10

The logarithmic expression

a1" title="log_{b}(\frac{1}{b^{-b}} )" alt="log_{b}(\frac{1}{b^{-b}} )" align="absmiddle" class="latex-formula"> where b > 0, is equal to
a) b^{-b}

b) b^{b}

c) -b

d) b
Mathematics
1 answer:
ivanzaharov [21]2 years ago
3 0

\log_b \left(\dfrac 1{b^{-b}}\right)\\\\=\log_b (b^{-b})^{-1}\\\\=\log_b  b^b\\\\=b \log_b b\\\\=b \cdot 1\\\\=b

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Answer:

407.22 foot is the boat from the base of the lighthouse

Step-by-step explanation:

Given the statement: An observer on top of a 50-foot tall lighthouse sees a boat at a 7° angle of depression.

Let x foot be the distance of the object(boat) from the base of the lighthouse

Angle of depression = 7^{\circ}

\angle CAB = \angle DCA = 7^{\circ}       [Alternate angle]

In triangle CAB:

To find AB = x foot.

Using tangent ratio:

\tan (\theta) = \frac{Opposite side}{Adjacent Base}

\tan (\angle CAB) = \frac{BC}{AB}

Here, BC = 50 foot and \angle CAB =7^{\circ}

then;

\tan (7^{\circ}) = \frac{50}{x}

or

x = \frac{50}{\tan 7^{\circ}}

x = \frac{50}{0.1227845609}

Simplify:

AB = x = 407.217321 foot

Therefore, the boat from the base of the light house is, 407.22'





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Answer:

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