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Gemiola [76]
3 years ago
15

HELP MEH PLEASE =) CLICK IF U DARE!

Mathematics
1 answer:
Lady bird [3.3K]3 years ago
8 0
Hmm well,
A) Talk
B)nice
C) want
D) that
the statement "do you want to talk to that nice salesperson" is much more biased. it is purposely making that particular salesperson seem better than the other in a way. hope this helped! :)
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PROVE THAT:<br><br>cos 20° - sin 20° = ​​ \sqrt{2}sin25°<br><br>​
Lemur [1.5K]

Answer:

See below.

Step-by-step explanation:

\cos(20)-\sin(20)=\sqrt{2}\sin(25)

First, use the co-function identity:

\sin(90-x)=\cos(x)

We can turn the second term into cosine:

\sin(20)=\sin(90-70)=\cos(70)

Substitute:

\cos(20)-\cos(70)=\sqrt{2}\sin(25)

Now, use the sum to product formulas. We will use the following:

\cos(x)-\cos(y)=-2\sin(\frac{x+y}{2})\sin(\frac{x-y}{2})

Substitute:

\cos(20)-\cos(70)=-2\sin(\frac{20+70}{2})\sin(\frac{20-70}{2})\\\cos(20)-\cos(70)  =-2\sin(45)\sin(-25)\\\cos(20)-\cos(70)=-2(\frac{\sqrt{2}}{2})\sin(-25)\\ \cos(20)-\cos(70)=-\sqrt{2}\sin(-25)

Use the even-odd identity:

\sin(-x)=-\sin(x)

Therefore:

\cos(20)-\cos(70)=-\sqrt{2}\sin(-25)\\\cos(20)-\cos(70)=-\sqrt{2}\cdot-\sin(25)\\\cos(20)-\cos(70)=\sqrt{2}\sin(25)

Replace the second term with the original term:

\cos(20)-\sin(20)=\sqrt{2}\sin(25)

Proof complete.

4 0
4 years ago
What is the interquartile range of this data set. 4,5,7,9,10,14,16,24
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Answer:

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Step-by-step explanation:

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3 years ago
Which relation is a function?
guapka [62]

Answer:

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Step-by-step explanation:


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3 years ago
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Step-by-step explanation:

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