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alex41 [277]
4 years ago
9

What is 15/8 as a mixed number

Mathematics
2 answers:
9966 [12]4 years ago
6 0
It would be equal to 1 7/8
romanna [79]4 years ago
3 0
1 and 3/4 would be the answer :)
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Camilla has three times as many points as Lynn. Lynn has five
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Use your understanding of the unit circle and trigonometric functions to find the values requested.
vfiekz [6]

Answer:

a) For this case we can use the fact that sin (\pi/3) = \frac{\sqrt{3}}{2}

And for this case since we ar einterested on -\frac{\pi}{3} and we know that the if we are below the y axis the sine would be negative then:

sin (-\pi/3) = -\frac{\sqrt{3}}{2}

b) From definition we can use the fact that tan x= \frac{sin x}{cos x} and we got this:

tan (5\pi/4) = \frac{sin(5\pi/4)}{cos(5\pi/4)}

We can use the notabl angle \pi/4 and we know that :

sin (\pi/4) = cos(\pi/4) = \frac{\sqrt{2}}{2}

Then we know that 5\pi/4 correspond to 225 degrees and that correspond to the III quadrant, and we know that the sine and cosine are negative on this quadrant. So then we have this:

tan (5\pi/4) = \frac{sin(5\pi/4)}{cos(5\pi/4)}= \frac{\frac{sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1

Step-by-step explanation:

For this case we can use the notable angls given on the picture attached.

Part a

For this case we can use the fact that sin (\pi/3) = \frac{\sqrt{3}}{2}

And for this case since we ar einterested on -\frac{\pi}{3} and we know that the if we are below the y axis the sine would be negative then:

sin (-\pi/3) = -\frac{\sqrt{3}}{2}

Part b

From definition we can use the fact that tan x= \frac{sin x}{cos x} and we got this:

tan (5\pi/4) = \frac{sin(5\pi/4)}{cos(5\pi/4)}

We can use the notabl angle \pi/4 and we know that :

sin (\pi/4) = cos(\pi/4) = \frac{\sqrt{2}}{2}

Then we know that 5\pi/4 correspond to 225 degrees and that correspond to the III quadrant, and we know that the sine and cosine are negative on this quadrant. So then we have this:

tan (5\pi/4) = \frac{sin(5\pi/4)}{cos(5\pi/4)}= \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1

6 0
3 years ago
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