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Ivahew [28]
3 years ago
13

Which set of ordered pairs does not represent a function? A. {(10, 1), (–2, –16), (–6, 7), (5, 1), (8, –16), (–11, 9)} B. {(–2,

–6), (–8, 13), (–7, –4), (–2, –8), (11, 8)} C. {(15, –3), (–6, 4), (–3, 0), (–1, 16)} D. {(6, 2), (–12, –16), (–6, 10), (20, 2)}
Mathematics
1 answer:
Gnoma [55]3 years ago
4 0

Answer:

B. {(–2, –6), (–8, 13), (–7, –4), (–2, –8), (11, 8)}

Step-by-step explanation:

A set of ordered pairs represents a function if for each input, there is only one output.

In this question:

In option B, for the input -2, there are two outputs(-6 and -8). This means that B is the correct answer.

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mario62 [17]

Answer:

zw=76 * (cos(\frac{3\pi}{16}) +isin(\frac{3\pi}{16}) )

Step-by-step explanation:

Given

z =38 (cos(\frac{\pi}{8})+ i sin(\frac{\pi}{8}) and

w = 2 (cos(\frac{\pi}{16} ) + i sin(\frac{\pi}{16})

Required

Determine zw

Given that:

x = v_1(cos(a) + isin(a)) and y = v_2(cos(b) + isin(b))

x * y = (v_1 * v_2)*((cos(a+b) + isin(a+b))

So, we have:

zw=(38 * 2) * (cos(\frac{\pi}{8} + \frac{\pi}{16}) + isin(\frac{\pi}{8} + \frac{\pi}{16}))

zw=76 * (cos(\frac{2\pi + \pi}{16})+isin(\frac{2\pi + \pi}{16}))

zw=76 * (cos(\frac{3\pi}{16}) +isin(\frac{3\pi}{16}) )

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3 years ago
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Answer:

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