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elena-14-01-66 [18.8K]
3 years ago
5

Solve this equation: 8(3x - 6) = 6(4x + 8) *

Mathematics
1 answer:
lana [24]3 years ago
5 0

Hey there!

8(3x - 6) = 6(4x + 8)

8(3x) + 8(-6) = 6(4x) + 6(8)

24x - 48 = 24x + 48

SUBTRACT 24x to BOTH SIDES

24x - 48 - 24x = 24x + 48 - 24

SIMPLIFY IT!

NEW EQUATION: -48 = 48

ADD 48 to BOTH SIDES

-48 + 48 = 48 + 48

NEW EQUATION: 0 = 96

Thus this makes your equation has

NO SOLUTION.

Therefore, your answer is:

Option A. Because this is a false

statement, the equation has no solution.

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

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Answer:

the maximum concentration of the antibiotic during the first 12 hours is 1.185 \mu g/mL at t= 2 hours.

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First, we differentiate C(t) with respect to t, to get,

\frac{d(C(t))}{dt} = 8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)})

Equating the first derivative to zero, we get,

\frac{d(C(t))}{dt} = 0\\\\8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0

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8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0\\\displaystyle\frac{e^{-0.4}}{e^{-0.6}} = \frac{0.6}{0.4}\\\\e^{0.2t} = 1.5\\\\t = \frac{ln(1.5)}{0.2}\\\\t \approx 2

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C(0) = 8(e^{(0)}-e^{(0)}) = 0

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C(2) = 8(e^{(-0.8)}-e^{(-1.2)}) = 1.185

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C(12) = 8(e^{(-4.8)}-e^{(-7.2)}) = 0.059

Thus, the maximum concentration of the antibiotic during the first 12 hours is 1.185 \mu g/mL at t= 2 hours.

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