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kodGreya [7K]
3 years ago
5

Can some0ne help me?

Mathematics
2 answers:
xxMikexx [17]3 years ago
6 0

g(x) = 3x + 2

y = 3x + 2

x = 3y + 2

3y = x - 2

y = x/3 - 2/3

inverse g(x) = (x - 2) / 3

g(x) = 4 - 5x

y = 4 - 5x

x = 4 - 5y

5y = 4 - x

y = 4/5 - x/5

inverse g(x) = (4 - x) / 5

PtichkaEL [24]3 years ago
4 0

Answer:

(x - 2)/3

(x - 4)/-5 or (-x + 4)/5

Step-by-step explanation:

this is an inverse function, and to solve an inverse function you would :

swap x and g(x) without bringing the x coefficient with it, just simply swap the variables. Then, solve for g(x), and that's it

the first question's answer is :

g(x) = 3x + 2

x = 3(g(x)) + 2

x - 2 = 3(g(x))

(x - 2)/3 = g(x)

the second one is:

g(x) = 4 - 5x

x = 4 - 5(g(x))

x - 4 = -5(g(x))

(x-4)/-5 = g(x)

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Step-by-step explanation:

Hello!

You have the following hypothesis:

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◆ COMPLEX NUMBERS ◆

given \: expression \: is \: - \\ \\ - 3 {i}^{4} + 2 {i}^{3} + 2 {i}^{2} + \sqrt{ - 9} \\ \\ we \: know \: that \: , \: \\ i = \sqrt{ - 1} \\ \\ {i}^{2} = - 1 \\ \\ {i}^{3 } = - i \\ \\ {i}^{4} = 1 \\ \\ \\ - 3 {i}^{4} + 2 {i}^{3} + 2 {i}^{2} + \sqrt{9} i \\ = - 3 {i}^{4} + 2 {i}^{3} + 2 {i}^{2} + 3i \\ \\ using \: the \: above \: properties \: of \: iota \: (i) \\ \\ = - 3(1) + 2( - i) + 2( - 1) + 3i \\ \\ = - 3 - 2i - 2 + 3i \\ \\ = (- 5 + i) \: \: \: ans.
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