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yulyashka [42]
3 years ago
12

What is the relationship between a and b

Mathematics
1 answer:
Mademuasel [1]3 years ago
4 0

Answer:

Vertical Angles

Step-by-step explanation:

Any opposite angles caused by two intersecting lines are vertical angles.

Any two adjacent angles caused by vertical lines or two angles that equal 180° are supplementary angles.

Two angles that make 90° in all are complementary angles.

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

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Answer:

\cos4( \theta)=Re(\cos4( \theta)+i\sin4( \theta)) \\=Re(\cos ( \theta)+i\sin ( \theta))^4 \\  =Re(cos^4( \theta)+4cos^3( \theta)isin( \theta)+6cos^2( \theta)i^2sin^2( \theta)+4cosi^3( \theta)sin^3( \theta)+i^4sin^4( \theta)) \\cos( 4\theta) =cos^4( \theta)-6cos^2( \theta)sin^2( \theta)+sin^4( \theta) \\ =(1-\sin^2( \theta))^2-6(1-\sin^2( \theta))\sin^2( \theta)+sin^4( \theta) \\ =  1-2\sin^2( \theta)+\sin^4( \theta)-6\sin^2( \theta)+6\sin^4( \theta)+\sin^4( \theta) \\ =8\sin^4( \theta)-8\sin^2( \theta)+1

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3 years ago
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Need help pleaseI was bad at math in school so lwant to learn
aleksklad [387]

The probability of an event is expressed as

Pr(\text{event) =}\frac{Total\text{ number of favourable/desired outcome}}{Tota\text{l number of possible outcome}}

Given:

\begin{gathered} \text{Red}\Rightarrow2 \\ \text{Green}\Rightarrow3 \\ \text{Blue}\Rightarrow2 \\ \Rightarrow Total\text{ number of balls = 2+3+2=7 balls} \end{gathered}

The probability of drwing two blue balls one after the other is expressed as

Pr(\text{blue)}\times Pr(blue)

For the first draw:

\begin{gathered} Pr(\text{blue) = }\frac{number\text{ of blue balls}}{total\text{ number of balls}} \\ =\frac{2}{7} \end{gathered}

For the second draw, we have only 1 blue ball left out of a total of 6 balls (since a blue ball with drawn earlier).

Thus,

\begin{gathered} Pr(\text{blue)}=\frac{number\text{ of blue balls left}}{total\text{ number of balls left}} \\ =\frac{1}{6} \end{gathered}

The probability of drawing two blue balls one after the other is evaluted as

\begin{gathered} \frac{1}{6}\times\frac{2}{7} \\ =\frac{1}{21} \end{gathered}

The probablity that none of the balls drawn is blue is evaluted as

\begin{gathered} 1-\frac{1}{21} \\ =\frac{20}{21} \end{gathered}

Hence, the probablity that none of the balls drawn is blue is evaluted as

\frac{20}{21}

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