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3 years ago

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How to square imaginary numbers

1 answer:

dalvyx [7]3 years ago

5 0

An imaginary number " i " is the squared root of -1, so whenever you square i it's like squaring a squared root. The squared root would cancel and you would be left with just the number under it, that is, the -1.
i ^ 2 = -1

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Please help asap, will mark Brainliest xoxo

IgorC [24]

Answer/Step-by-step explanation:

Given, $b(x) = (\frac{6}{7})^{x}$

The table for the function are:

When x = -2

$b(-2) = (\frac{6}{7})^{-2}$

$b(-2) = \frac{1}{(\frac{6}{7})^{2}}$

$b(-2) = \frac{1}{(\frac{36}{49})}$

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$b(-2) = \frac{49}{36}$

When x = -1

$b(-1) = (\frac{6}{7})^{-1}$

$b(-1) = \frac{1}{(\frac{6}{7})}$

$b(-1) = 1*\frac{7}{6}$

$b(-2) = \frac{7}{6}$

When x = 0

$b(0) = (\frac{6}{7})^{0}$

$b(0) = \frac{6^0}{7^0}$

$b(0) = \frac{1}{1}$

$b(0) = 1$

When x = 1

$b(1) = (\frac{6}{7})^{1}$

$b(1) = \frac{6}{7}$

When x = 2

$b(2) = (\frac{6}{7})^{2}$

$b(2) = \frac{6^2}{7^2}$

$b(2) = \frac{36}{49}$

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Can somebody help me with this problem

kati45 [8]

First substitute y=5x+7 so that would turn into 5x+7=6x

Then, isolate/solve for x for 5x + 7 = 6x, in this problem x=7

Substitute x=7 into y=5x+7

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