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3 years ago

Sketch one cycle of the cosine function y= cos 2θ

1 answer:

7 0

Based on function transformations, the graph of y= cos(2θ), is the same as the graph of <span>y= cos(θ), with a component B of 2, in which case, it means the period is the only thing that differs

well the period for this function will then be </span> $\bf \cfrac{2\pi }{B}\iff \cfrac{2\pi }{2}\implies \pi$

so, the graph of is, the same as y= cos(θ), just shrunk by half horizontally

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How do I find the distance between (-2,5) and (22,12)

jok3333 [9.3K]

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-2}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{22}~,~\stackrel{y_2}{12})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[22-(-2)]^2+[12-5]^2}\implies d=\sqrt{(22+2)^2+(12-5)^2} \\\\\\ d=\sqrt{24^2+7^2}\implies d=\sqrt{625}\implies d=25$

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3 years ago

Read 2 more answers

Which of the following is not one of the 8th roots of unity?

Anika [276]

Answer:

1+i

Step-by-step explanation:

To find the 8th roots of unity, you have to find the trigonometric form of unity.

1. Since $z=1=1+0\cdot i,$ then

$Rez=1,\\ \\Im z=0$

and

$|z|=\sqrt{1^2+0^2}=1,\\ \\\\\cos\varphi =\dfrac{Rez}{|z|}=\dfrac{1}{1}=1,\\ \\\sin\varphi =\dfrac{Imz}{|z|}=\dfrac{0}{1}=0.$

This gives you $\varphi=0.$

Thus,

$z=1\cdot(\cos 0+i\sin 0).$

2. The 8th roots can be calculated using following formula:

$\sqrt[8]{z}=\{\sqrt[8]{|z|} (\cos\dfrac{\varphi+2\pi k}{8}+i\sin \dfrac{\varphi+2\pi k}{8}), k=0,\ 1,\dots,7\}.$

Now

at k=0, $z_0=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 0}{8}+i\sin \dfrac{0+2\pi \cdot 0}{8})=1\cdot (1+0\cdot i)=1;$

at k=1, $z_1=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 1}{8}+i\sin \dfrac{0+2\pi \cdot 1}{8})=1\cdot (\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};$

at k=2, $z_2=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 2}{8}+i\sin \dfrac{0+2\pi \cdot 2}{8})=1\cdot (0+1\cdot i)=i;$

at k=3, $z_3=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 3}{8}+i\sin \dfrac{0+2\pi \cdot 3}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};$

at k=4, $z_4=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 4}{8}+i\sin \dfrac{0+2\pi \cdot 4}{8})=1\cdot (-1+0\cdot i)=-1;$

at k=5, $z_5=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 5}{8}+i\sin \dfrac{0+2\pi \cdot 5}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};$

at k=6, $z_6=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 6}{8}+i\sin \dfrac{0+2\pi \cdot 6}{8})=1\cdot (0-1\cdot i)=-i;$

at k=7, $z_7=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 7}{8}+i\sin \dfrac{0+2\pi \cdot 7}{8})=1\cdot (\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};$

The 8th roots are

$\{1,\ \dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ i, -\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ -1, -\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2},\ -i,\ \dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2}\}.$

Option C is icncorrect.

5 0

3 years ago

Write an equation in point-slope form of the line through the point (−1, 2) and with slope m = 2.

Katen [24]

Answer:

D. y - 2 = 2(x + 1)

Step-by-step explanation:

y - 2 = 2 (x - (-1))

y - 2 = 2 (x + 1)

<h3>#CMIIW</h3>

3 0

3 years ago

Determine whether y varies directly with x. If so, find the constant of variation and write the function rule.

Alexeev081 [22]

Yes they are, the constant variation is that for every 3 added to the x side, 1 is added to the y side. I would really appreciate the brainliest mark thanks!

4 0

3 years ago

10. Subtract the<br>fractions.<br>20 1 / 2 - 15 1/2 =<br>*Bonus: Reduce the<br>fraction.

BartSMP [9]

Answer: 5

Step-by-step explanation:

$20\frac{1}{2}-15\frac{1}{2}$

$20-15(\frac{1}{2}-\frac{1}{2})$

3 0

3 years ago