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3 years ago

Endpoint: (1,7), midpoint: (-10,10)

Mathematics

1 answer:

algol [13]3 years ago

5 0

The answer is (-21, 13) for The second endpoint.

Let's start by calling the known endpoint L and the unknown K. We'll call the midpoint M. In order to find this, we must first note that to find a midpoint we need to take the average of the endpoints. To do this we add them together and then divide by 2. So, using that, we can write a formula and solve for each part of the k coordinates. We'll start with just x values.

(Kx + Lx)/2 = Mx

(Kx + 1)/2 = -10

Kx + 1 = -20

Kx = -21

And now we do the same thing for y values

(Ky + Ly)/2 = My

(Ky + 7)/2 = 10

Ky + 7 = 20

Ky = 13

This gives us the final point of (-21, 13)

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3 years ago

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2 years ago

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kicyunya [14]

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Step-by-step explanation:

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3 years ago

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

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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.

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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2 years ago