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densk [106]
3 years ago
15

Kyra is framing a square painting with side lengths of (x + 6) inches.

Mathematics
1 answer:
mariarad [96]3 years ago
3 0

Answer:

me dont know man why yall givin me this question

Step-by-step explanation:

ahhhhhhhhhhhhhhhhh

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2 years ago
9. Given the point (6,-8) values of the six trig function.
lozanna [386]

Answer:

Here's what I get.

Step-by-step explanation:

9. (6, -8)

The reference angle θ is in the fourth quadrant.

∆AOB is a right triangle.

OB² = OA² + AB² = 6² + (-8)² = 36 + 64 = 100

OB = √100 = 10  

\sin \theta = \dfrac{-8}{10} = -\dfrac{4}{5}\\\\\cos \theta =\dfrac{6}{10} = \dfrac{3}{5}\\\\\tan \theta = \dfrac{-8}{6} = -\dfrac{4}{3}\\\\\csc \theta = \dfrac{10}{-8} = -\dfrac{5}{4}\\\\\sec \theta = \dfrac{10}{6} = \dfrac{5}{3}\\\\\cot \theta = \dfrac{6}{-8} = -\dfrac{3}{4}

10. cot θ = -(√3)/2

The reference angle θ is in the second quadrant.

∆AOB is a right triangle.

OB² = OA² + AB² = (-√3)² + (2)² = 3 + 4 = 7

OB = √7

\sin \theta = \dfrac{2}{\sqrt{7}} = \dfrac{2\sqrt{7}}{7}\\\\\cos \theta = \dfrac{-\sqrt{3}}{\sqrt{7}} = -\dfrac{\sqrt{21}}{7}\\\\\tan \theta = \dfrac{2}{-\sqrt{3}} = -\dfrac{2\sqrt{3}}{3}\\\\\csc \theta = \dfrac{\sqrt{7}}{2} \\\\\sec \theta = \dfrac{\sqrt{7}}{-\sqrt{3}} = -\dfrac{\sqrt{21}}{3}\\\\\cot \theta = -\dfrac{\sqrt{3}}{2}

3 0
2 years ago
What is tan 30°?<br> | sa"<br> зо“<br> -<br> -<br> оооооо<br> - 5<br> 3<br> ~|
Mnenie [13.5K]

Answer:

E

Step-by-step explanation:

4 0
3 years ago
All four B points are equidistant from points A and C. Are they all midpoints of AC? Justify your answer.
Rina8888 [55]

Answer:

No

Step-by-step explanation:

For a point to be the midpoint of a line segment, it must bisect it into two equal segments and be on the line segment (hence, colinear with the endpoints). All four B points are equidistant from points A and C, but aren’t colinear with A and C. Therefore, they aren’t all midpoints of line segment AC.

I hope this helps! :)

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2 years ago
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