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DochEvi [55]
3 years ago
9

A student tries out for the track team and runs a lap in 45 seconds. The coach tells her she will have to cut her time by 50% to

make the team. How fast must she run a lap to make the team?
Mathematics
2 answers:
Fiesta28 [93]3 years ago
8 0
Cutting her time by 50% means reducing her time by half. Half of 45 seconds is 22.5 seconds. Therefore, to make the team, she must run a lap in 22.5 seconds.
Scrat [10]3 years ago
5 0

Answer:

Cutting her time by 50% means reducing her time by half. Half of 45 seconds is 22.5 seconds. Therefore, to make the team, she must run a lap in 22.5 seconds.

Step-by-step explanation:

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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}

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A

Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

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