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eduard
2 years ago
10

Find the exact value.

Mathematics
1 answer:
levacccp [35]2 years ago
4 0

\text{Given that,}\\\\\cot A = -\dfrac{9}{\sqrt{40}}\\\\\\\implies \tan A = -\dfrac{\sqrt{40}}9\\\\\\\implies \tan^2 A = \dfrac{40}{81}\\\\\\\implies \sec^2 A -1 = \dfrac{40}{81}\\\\\implies \sec^2 A = 1 + \dfrac{40}{81}\\\\\implies \sec^2 A = \dfrac{121}{81}\\\\\implies \sec A = \pm \sqrt{\dfrac{121}{81}}= \pm\dfrac{11}{9}\\\\\\\text{Since the angle A is in quadrant II, the value of sec A will be negative.}\\\\\\\text{Hence,}~ \sec A=-\dfrac{11}{9}

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MArishka [77]

Answer:

x = 28

Step-by-step explanation:

Given that lines AB and CD are straight lines that intersects at O, it follows that the pair of opposite vertical angles formed are congruent.

Thus,

<AOD = <BOC

<AOD = 152°

<BOC = 3x + x + (x + 12) (angle addition postulate)

<BOC = 5x + 12

Since <AOD = <BOC, therefore,

152° = 5x + 12 (substitution)

152 - 12 = 5x (subtraction property of equality)

140 = 5x

140/5 = x (division property of equality)

28 = x

x = 28

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3 years ago
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Anna007 [38]

Answer:

b.

Step-by-step explanation:

We have to look at sign changes in f(x) to determine the possible positive real roots.

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There is only one sign change here, between the -8x and the +4.  So that means there is only 1 possible real positive root.

Now we have to look at sign changes in f(-x) to determine the possible negative real roots.

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If we have 1 positive and 1 negative, we have to have 2 imaginary

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Keep in mind that the total number or roots--positive, negative, imaginary--have to add up to equal the degree of the polynomial.  This is a 4th degree polynomial, so we will have 4 roots.

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