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Vadim26 [7]
3 years ago
7

"absmiddle" class="latex-formula">
Mathematics
1 answer:
ICE Princess25 [194]3 years ago
4 0

Answer:

  x = 162

Step-by-step explanation:

The equation editor lets you properly write exponents, so we have to assume that your equation is supposed to be equivalent to ...

  27 = x/6

  27·6 = x = 162 . . . . multiply by 6

_____

On the off chance you intend ...

27=\left(\dfrac{1}{3}\right)^{\dfrac{x}{2}}

you can solve this by taking logarithms and dividing by the coefficient of x. Using base 3 for the logarithm may be easiest.

\log_3{(27)}=\dfrac{x}{2}\log_3{(3^{-1})}\\\\3=-\dfrac{x}{2} \qquad\text{simplify}\\\\-6=x \qquad\text{multiply by -2}

This version of the equation has solution x = -6.

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PLZ RESPOND ASAP HELPPPP
Ronch [10]

The question is the thing that needs help. What does that hanging "a leading 2" mean? And rational real coefficients? Obviously if they're rational they're gonna be real, but really.


I would say "a leading 2" means the highest power of x has a coefficient to two, which is none of the above.


The last two are of degree two, which is the lowest degree. They both have integer coefficients, so are necessarily real and rational as well.


Neither has a leading 2 as far as I can tell. The last one is monic (a leading coefficient of 1). I like monic polynomials so I'd pick that one, but that doesn't make it right.


None of the above



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3 years ago
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erma4kov [3.2K]

Answer:

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Step-by-step explanation:

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it is corrected
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true

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