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s2008m [1.1K]
3 years ago
9

Please Help me To solve this Problem

Mathematics
1 answer:
mart [117]3 years ago
3 0

You're given that φ is an angle that terminates in the third quadrant (III). This means that both cos(φ) and sin(φ), and thus sec(φ) and csc(φ), are negative.

Recall the Pythagorean identity,

cos²(φ) + sin²(φ) = 1

Multiply the equation uniformly by 1/cos²(φ),

cos²(φ)/cos²(φ) + sin²(φ)/cos²(φ) = 1/cos²(φ)

1 + tan²(φ) = sec²(φ)

Solve for sec(φ) :

sec(φ) = - √(1 + tan²(φ))

Given that cot(φ) = 1/4, we have tan(φ) = 1/cot(φ) = 1/(1/4) = 4. Then

sec(φ) = - √(1 + 4²) = -√17

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4 meters

Step-by-step explanation:

Given a quadratic equation in which the coefficient of x^2 is negative, the parabola opens up and has a maximum point. This maximum point occurs at the line of symmetry.

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Step 2: Find the value of y at the point of symmetry

That is, we substitute x obtained above into the y and solve.

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Find the equation of all tangent lines having slope of -1 that are tangent to the curve y=(9)/(x+1)
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First tangent line:

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Second tangent line:

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Notice: slope of -1 means that both f'(x_1), \ f'(x_2) are equal to -1, so f'(x_1)=-1 \ and \ f'(x_2)=-1


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