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MariettaO [177]
2 years ago
12

Find the tangent of ∠I.

Mathematics
1 answer:
snow_tiger [21]2 years ago
8 0

The simplified, rationalized form is tan I = √70/5

<h3>SOH CAH TOA identity</h3>

From the given diagram, we are given the following parameters

  • Adjacent to m<I = IG
  • Hypotenuse = √95

According to the SOH CAH TOA identity;

tan m<I = opp/adj

tanm<I = Determine the opposite side using the pythagoras theorem:

GH² = (√95)² - 5²

GH² = 95 - 25

GH² = 70

GH = √70

Determine the value of tanm<I

tan I = √70/5

Hence the simplified, rationalized form is tan I = √70/5

Learn more on SOH CAH TOA here: brainly.com/question/20734777

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Choice A is correct.
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If y = 3x^3 - 2x and dx/dt equals 3, find dy/dt when x = -2. Give only the numerical answer. For example, if dy, dt = 4, type on
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Answer:

 \frac{dy}{dt}  = 102

Step-by-step explanation:

<u>step(i)</u>

Given function y = 3 x³ - 2 x  ... (i)

Differentiating equation (i) with respective to 'x'

    \frac{dy}{dx} = 3 (3 x^{2} ) - 2(1)

Given x = -2

(\frac{dy}{dx} ) x_{=-2} = 3 (3 (-2)^{2} ) - 2(1)

(\frac{dy}{dx} ) x_{=-2} = 36 -2 =34

<u><em>Step(ii):-</em></u>

<u><em>we know that </em></u>

                     \frac{dy}{dx} = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }

Given    \frac{dx}{dt} = 3  and \frac{dy}{dx} = 34

                  \frac{dy}{dx} = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }

substitute values    \frac{dx}{dt} = 3  and \frac{dy}{dx} = 34  

           ⇒        34 = \frac{\frac{dy}{dt} }{3 }

cross multiplication , we get

                      \frac{dy}{dt} = 34( 3 ) = 102

<u><em>Final answer</em></u>:-

 \frac{dy}{dt} = 34( 3 ) = 102

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