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irga5000 [103]
3 years ago
12

Tamara says that raising the number i to any integer power results in either -1 or 1 as the result, since i^2 = -1. Do you agree

d or disagree with Tamera? Explain
Mathematics
1 answer:
Roman55 [17]3 years ago
5 0
I agree only if you have even powers -- even negative ones.

1/i^2 = 1/-1 = - 1
i^0 also gives 1 So far no problem.

It is when you consider the odd numbers that you don't get 1 or -1 
You get either -i or i
i^(4n + 1) =  i
i^(4n - 1) = -i

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20 miles $80

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Suppose sin(A) = 1/4. use the trig identity sin^2(A)+cos^2(A)=1 to find cos(A) in quadrant II. around to ten-thousandth.
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In quadrant II, \cos(A) will be negative. So

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8 0
2 years ago
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jeyben [28]
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3 0
3 years ago
URGENT!!
Lapatulllka [165]

Answer:

f[g(4)] = 4

Step-by-step explanation:

Given table:

\begin{array}{| c | c | c | c | c | c |}\cline{1-6} x & -6 & -4 & 1 & 3 & 4\\\cline{1-6} f(x) & 4 & -1 & -6 & 1 & 3 \\\cline{1-6} g(x) & 1 & 4 & 3 & -4 & -6 \\\cline{1-6}\end{array}

f[g(4)] is a composite function.

When calculating <u>composite functions</u>, always work from inside the brackets out.

Begin with g(4):  g(4) is the value of function g(x) when x = 4.

From inspection of the given table, g(4) = -6

Therefore, f[g(4)] = f(-6)

f(-6) is the value of function f(x) when x = -6.

From inspection of the given table, f(-6) = 4

Therefore, f[g(4)] = 4

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2 years ago
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