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Ostrovityanka [42]
3 years ago
7

DUE IN 30 MINUTES HELP NEEDED ASAP!!!!

Mathematics
2 answers:
Anna71 [15]3 years ago
6 0
It would be R1 R2 R3 R4 R5 & R6
statuscvo [17]3 years ago
5 0
It’s R1 r2 r3 r4 r5 r6
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What are the solutions of the equation x^4 + 6x^2 + 5 = 0? Use u substitution to solve.
Radda [10]

Answer:

second option

Step-by-step explanation:

Given

x^{4} + 6x² + 5 = 0

let u = x², then

u² + 6u + 5 = 0 ← in standard form

(u + 1)(u + 5) = 0 ← in factored form

Equate each factor to zero and solve for u

u + 1 = 0 ⇒ u = - 1

u + 5 = 0 ⇒ u = - 5

Change u back into terms of x, that is

x² = - 1 ( take the square root of both sides )

x = ± \sqrt{-1} = ± i ( noting that \sqrt{-1} = i ), and

x² = - 5 ( take the square root of both sides )

x = ± \sqrt{-5} = ± \sqrt{5(-1)} = ± \sqrt{5} × \sqrt{-1} = ± i\sqrt{5}

Solutions are x = ± i and x = ± i\sqrt{5}

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2 years ago
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Mariulka [41]
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The least common multiple is 3

Hope this helps!
4 0
3 years ago
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prohojiy [21]

Step-by-step explanation:

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