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3 years ago

Can y’all help me on this ? I kinda forgot how to do this-

Mathematics

1 answer:

Mashutka [201]3 years ago

7 0

Answer: 62

Do you need an explanation?

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Luke had $9. He had to spend 3/4 of it for a birthday gift. How much cash money did<br> he spend?

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He spent $2.25.

Step-by-step explanation:

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Now subtract $6.75 from $9.

We get $2.25

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3 years ago

Use the unit circle to evaluate these expressions:

sergeinik [125]

Answer:

a) We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to $2\pi$ :

$17 \pi/4 - 2\pi = \frac{9\pi}{4} -2\pi = \pi/4$

For this case we know that $sin (\pi/4) = \frac{\sqrt{2}}{2}$

So then $sin(\frac{17 \pi}{4}) = \frac{\sqrt{2}}{2}$

b) We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to $2\pi$ :

$19 \pi/6 - 2\pi = \frac{7\pi}{6}$

For this case we know that $cos (\pi/6) = \frac{\sqrt{3}}{2}$

And we know that $\frac{7\pi}{6}$ is on the III quadrant since is equivalent to 210 degrees. And on the III quadrant the cosine is negative. So then $cos(\frac{19 \pi}{6}) = -\frac{\sqrt{3}}{2}$

c) For this case that any factor of $\pi$ the sin function is equal to 0.

So from definition of tan we have this:

$tan (450\pi) = \frac{sin(450 \pi)}{cos(450 \pi)}= \frac{0}{cos(450\pi)}= 0$

Step-by-step explanation:

a. sin (17pi / 4 )

We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to $2\pi$ :

$17 \pi/4 - 2\pi = \frac{9\pi}{4} -2\pi = \pi/4$

For this case we know that $sin (\pi/4) = \frac{\sqrt{2}}{2}$

So then $sin(\frac{17 \pi}{4}) = \frac{\sqrt{2}}{2}$

b. cos (19pi / 6 )

We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to $2\pi$ :

$19 \pi/6 - 2\pi = \frac{7\pi}{6}$

For this case we know that $cos (\pi/6) = \frac{\sqrt{3}}{2}$

And we know that $\frac{7\pi}{6}$ is on the III quadrant since is equivalent to 210 degrees. And on the III quadrant the cosine is negative. So then $cos(\frac{19 \pi}{6}) = -\frac{\sqrt{3}}{2}$

c. tan(450pi)

For this case that any factor of $\pi$ the sin function is equal to 0.

So from definition of tan we have this:

$tan (450\pi) = \frac{sin(450 \pi)}{cos(450 \pi)}= \frac{0}{cos(450\pi)}= 0$

4 0

3 years ago

HELPPPPPP PLEASEEEEEEE

m_a_m_a [10]

Ok I’ll bet it will work for me and you

7 0

3 years ago