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Zanzabum
3 years ago
9

Define/ Describe "Tidal Bore. How does it apply to the Chesapeake Bay?

Geography
1 answer:
sergiy2304 [10]3 years ago
6 0

Answer:

A tidal bore is a rare natural phenomenon in which an incoming tide creates a wave of water that travels up along a river or a narrow bay causing water to flow against the river's current. Tidal bores occur in relatively few locations worldwide.

Tidal level in the Chesapeake Bay is not affected by moon phases. During the full and new moons, the high tidal level should be the highest. During 1st and 3rd quarter moons, the high tides should be the lowest.

Explanation:

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Because it sits on boundary.

Explanation:

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3 years ago
Suppose 0 is an angle in the standard position whose terminal side is in Quadrant IV and cot0= -6/7 Find the exact values of the
Sergio039 [100]
<h2>\sin \theta=\dfrac{p}{h} =\dfrac{-7}{\sqrt{85} }, \cos \theta=\dfrac{b}{h} =\dfrac{6}{\sqrt{85}}, \tan \theta=\dfrac{p}{b} =\dfrac{-7}{6},</h2><h2>\sec \theta=\dfrac{h}{b} =\dfrac{\sqrt{85} }{6} and \csc \theta=\dfrac{h}{p} =\dfrac{-\sqrt{85}}{ 7}</h2>

Explanation:

Given,

\cot \theta= \dfrac{-6}{7}

To find, the exact values of the five remaining trigonometric functions of \theta = ?

We know that,

\cot \theta= \dfrac{-6}{7}=\dfrac{b}{p}

Where, b = base and p = perpendicular

By Pythagoras's theorem,

Hypotaneous, h=\sqrt{p^2+b^2}

=\sqrt{(-6)^2+7^2}=\sqrt{36+49} =\sqrt{85}

In IVth quadrant,

\cot \theta and \sec \theta are positive.

\sin \theta=\dfrac{p}{h} =\dfrac{-7}{\sqrt{85} }, \cos \theta=\dfrac{b}{h} =\dfrac{6}{\sqrt{85}}, \tan \theta=\dfrac{p}{b} =\dfrac{-7}{6},

\sec \theta=\dfrac{h}{b} =\dfrac{\sqrt{85} }{6} and \csc \theta=\dfrac{h}{p} =\dfrac{-\sqrt{85}}{ 7}

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