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Elodia [21]
2 years ago
6

Point Q is plotted on the coordinate grid. Point P is at (20, −30). Point R is vertically above point Q. It is at the same dista

nce from point Q as point P is from point Q. Which of these shows the coordinates of point R and its distance from point Q?
On a coordinate grid from negative 50 to positive 50 in increments of 10, Point Q is plotted at the ordered pair negative 40, negative 30.

Point R is at (−40, 30), a distance of 60 units from point Q
Point R is at (−40, −10), a distance of 20 units from point Q
Point R is at (−40, −30), a distance of 60 units from point Q
Point R is at (−40, 10), a distance of 20 units from point Q
Mathematics
1 answer:
Elina [12.6K]2 years ago
4 0

Answer:

Point R is at (−40, −10), a distance of 20 units from point Q

Step-by-step explanation:

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Answer:

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3 years ago
Find out the answers to this
meriva

Answer:

\tan \theta = - \frac{1}{5} = - 0.2

\cos \theta = 0.98

\sin \theta = - 0.196

Step-by-step explanation:

It is given that \cot \theta = - 5 and \theta is in the fourth quadrant.

So, only \cos \theta will have positive value and \sin \theta, \tan \theta will have negative value.

Now, \cot \theta = - 5

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We know, that \sec^{2} \theta - \tan^{2} \theta = 1

⇒ \sec \theta = \sqrt{1 + \tan^{2} \theta } = \sqrt{1 + (- \frac{1}{5} )^{2} } = 1.019

{Since, \cos \theta is positive then \sec \theta will also be positive}

⇒ \cos \theta = \frac{1}{\sec \theta} = \frac{1}{1.0198} = 0.98 (Answer)

We know, that \csc^{2} \theta - \cot^{2} \theta = 1

⇒ \csc \theta = - \sqrt{1 + \cot^{2} \theta } = - \sqrt{1 + (- 5 )^{2} } = - 5.099

{Since, \sin \theta is negative then \csc \theta will also be negative}

⇒ \sin \theta = \frac{1}{\csc \theta} = \frac{1}{- 5.099} = - 0.196 (Answer)

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Answer:

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Step-by-step explanation:

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