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4 years ago

Hopefully this is enough for the answer

Mathematics

1 answer:

Tomtit [17]4 years ago

6 0

It is the third one.

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A function is represented by these ordered pairs.

Crazy boy [7]

answer: 4

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4 0

4 years ago

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What is the slope of the line that passes through the points (-2, 7) and (2, -5)?

amm1812

Answer:-3

Step-by-step explanation:

6 0

3 years ago

(sqrt3-sqrt3i)^4

Ludmilka [50]

The increasing order of the complex numbers is (√2 - i)⁶ < (√2 - √2i)⁸ = (√3 - i)⁶ = (-1 + √3i)¹² < (√3 - √3i)⁴.

<h3>Absolute values of the complex numbers</h3>

The absolute values of the complex numbers are determined as follows;

(sqrt3-sqrt3i)^4 = (√3 - √3i)⁴

$|z| = \sqrt{(\sqrt{3} )^2 + (\sqrt{3 }\times1 )^2} } \\\\|z| = \sqrt{6}$

(-1+sqrt3i)^12 = (-1 + √3i)¹²

$|z| = \sqrt{(-1)^2 + (\sqrt{3)^2} } \\\\|z| = \sqrt{4} \\\\|z| = 2$

(sqrt 3-i)^6 = (√3 - i)⁶

$|z| = \sqrt{(\sqrt{3})^2 + (-1)^2 } \\\\|z| = \sqrt{4} \\\\|z| = 2$

(sqrt2-sqrt2i)^8 = (√2 - √2i)⁸

$|z| = \sqrt{(\sqrt{2} )^2 + (\sqrt{2})^2 } \\\\|z| = 2$

(sqrt2-i)^6 = (√2 - i)⁶

$|z| = \sqrt{(\sqrt{2})^2 + (-1)^2} } \\\\|z| = \sqrt{3}$

Increasing order of the complex numbers;

(√2 - i)⁶ < (√2 - √2i)⁸ = (√3 - i)⁶ = (-1 + √3i)¹² < (√3 - √3i)⁴.

Learn more about complex numbers here: brainly.com/question/10662770

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3 0

3 years ago

Staci has seven more cats than Taylor. write an expression that illustrates how many cats Staci has. Use x to represent the numb

laila [671]

X + 7 is your answer

4 0

3 years ago

Read 2 more answers

If sin theta = (4)/(7), theta in quadrant II, find the exact value of (a) cos theta (b) sin (theta + (pi) / (6) ) (c) cos (the

EleoNora [17]

Answer:

a) $\cos(\theta) = \frac{\sqrt[]{33}}{7}$

b) $\sin(\theta + \frac{\pi}{6})\frac{-3\sqrt[]{11}+4}{14}$

c) $\cos(\theta-\pi)=\frac{\sqrt[]{33}}{7}$

d) $\tan(\theta + \frac{\pi}{4}) = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}$

Step-by-step explanation:

We will use the following trigonometric identities

$\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)$

$\cos(\alpha+\beta) = \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$ $\tan(\alpha+\beta) = \frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$ .

Recall that given a right triangle, the sin(theta) is defined by opposite side/hypotenuse. Since we know that the angle is in quadrant 2, we know that x should be a negative number. We will use pythagoras theorem to find out the value of x. We have that

$x^2+4^2 = 7 ^2$

which implies that $x=-\sqrt[]{49-16} = -\sqrt[]{33}$ . Recall that cos(theta) is defined by adjacent side/hypotenuse. So, we know that the hypotenuse is 7, then

$\cos(\theta) = \frac{-\sqrt[]{33}}{7}$

b)Recall that $\sin(\frac{\pi}{6}) =\frac{1}{2} , \cos(\frac{\pi}{6}) = \frac{\sqrt[]{3}}{2}$ , then using the identity from above, we have that

$\sin(\theta + \frac{\pi}{6}) = \sin(\theta)\cos(\frac{\pi}{6})+\cos(\alpha)\sin(\frac{\pi}{6}) = \frac{4}{7}\frac{1}{2}-\frac{\sqrt[]{33}}{7}\frac{\sqrt[]{3}}{2} = \frac{-3\sqrt[]{11}+4}{14}$

c) Recall that $\sin(\pi)=0, \cos(\pi)=-1$ . Then,

$\cos(\theta-\pi)=\cos(\theta)\cos(\pi)+\sin(\theta)\sin(\pi) = \frac{-\sqrt[]{33}}{7}\cdot(-1) + 0 = \frac{\sqrt[]{33}}{7}$

d) Recall that $\tan(\frac{\pi}{4}) = 1$ and $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}=\frac{-4}{\sqrt[]{33}}$ . Then

$\tan(\theta+\frac{\pi}{4}) = \frac{\tan(\theta)+\tan(\frac{\pi}{4})}{1-\tan(\theta)\tan(\frac{\pi}{4})} = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}$

5 0

3 years ago