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3 years ago

the statement cot theta=12/5 sec theta= -13/5 and the terminal point determined by theta is in quadrant 4

Mathematics

2 answers:

hichkok12 [17]3 years ago

8 0

<h2>Answer:</h2>

The statement cot theta=12/5 sec theta= -13/5 and the terminal point determined by theta is in quadrant 4 is:

FALSE STATEMENT

<h2>Step-by-step explanation:</h2>

We are given an angle theta(θ) such that at that angle the value of two trignometric function are given as follows:

$\cot \theta=\dfrac{12}{5}\\\\and\\\\\\\sec \theta=\dfrac{-13}{5}$

and the terminal point determined by theta is in quadrant 4.

As we know that in fourth quadrant both the cosine as well as secant function is positive whereas the other trignometric function are negative but here the secant function is given negative which is a contradiction to our statement.

Hence, the given statement is a FALSE STATEMENT.

Sonbull [250]3 years ago

7 0

<span>cannot be true because secant theta is greater than zero in quadrant 4</span>

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The critical numbers/values are x = 0, 4/7, 2

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This is a doozy; no wonder you have it up here for help!

The critical numbers of a function are found where the derivative of the function is equal to 0. To find these numbers, you have to factor the deriative or simply solve it for 0. This one is especially difficult since it involves rational exponents that have to be factored. But this is fun, so let's get to it.

First off, I am assuming that the function is

$f(x)=x^{\frac{4}{5}}*(x-2)^2$ which involves using the product rule to find the derivative.

That derivative is

$f'(x)=x^{\frac{4}{5}}*2(x-2)+\frac{4}{5}x^{-\frac{1}{5}}(x-2)^2$ which simplifies down to

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$f'(x)=2x^{\frac{9}{5}}-4x^{\frac{4}{5}}+\frac{4}{5}x^{\frac{9}{5}}-\frac{16}{5}x^{\frac{4}{5}}+\frac{16}{5}x^{-\frac{1}{5}}$

Let's get everything over the common denominator of 5 so we can easily add and subtract like terms:

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Combining like terms gives us

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Those are the critical numbers/values for that function. This indicates where there is a max value or a min value.

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