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2 years ago

Solve for x A)3.0 B)3.25 C)3.5 D)3.75

Mathematics

1 answer:

tatuchka [14]2 years ago

3 0

Answer:

Answer: D) 3.75

Step-by-step explanation:

According to the Insersecting chords theorem, when you multiply the lengths of the segments of one of the chords in a circle, the product obtained is equal to the product of the segments of the other chord.

Based on this, you can find the value of "x" with:

Solve for "x" (Applying the Division property of equality, you can divide both sides of the equation by 4). Then:

This value matches with the option D.

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Step-by-step explanation:

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2 years ago

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Answer:

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Step-by-step explanation:

NOTE: I am kind of confused about the way this question is formatted, so I tried to answer everything I could decipher.

1 a.) –2t + 5t – 2 – t → 2t – 2

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Step-by-step explanation:

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2 years ago

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Step-by-step explanation:

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3 years ago

Read 2 more answers

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modelled by the function C(t)=8(e

Alexxx [7]

Answer:

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

Step-by-step explanation:

We are given the following information:

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modeled by the function where the time t is measured in hours and C is measured in $\mu g/mL$

$C(t) = 8(e^{(-0.4t)}-e^{(-0.6t)})$

Thus, we are given the time interval [0,12] for t.

We can apply the first derivative test, to know the absolute maximum value because we have a closed interval for t.
The first derivative test focusing on a particular point. If the function switches or changes from increasing to decreasing at the point, then the function will achieve a highest value at that point.

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,

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

Solving, we get,

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

At t = 0

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

At t = 2

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

At t = 12

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

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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