The question I need is question #13. Please provide a step by step answer. Thank You

Mathematics

1 answer:

Jet001 [13]3 years ago

4 0

Answer:

Step-by-step explanation:

We have plane 1 flying SW for 4 hours at a rate of 415 mph. The distance he covers using the d = rt formula for distance, is 415(4) = 160 miles.

We also have plane 2 flying directly east (along the x-axis) for 4 hours at 327 mph. The distance he covers using the d = rt formula for distance, is 327(4) = 1308 miles. The angle in between them at this point is 135 degrees, and what we need to find is the length of the vector connecting the 2 planes. IF this was right triangle trig that distance would be the hypotenuse and we could solve for it using Pythagorean's Theorem. BUT it is NOT a right triangle, so we have to find some other means with which to solve for that length. We will use the Law of Cosines to do this.

$?^2=1660^2+1308^2-2(1660)(1308)cos(135)$ which simplifies a bit to

$?^2=2755600+1712864-(-3070653.624)$

If you add all of that together, you'll get

$?^2=7537117.624$ and you'll take the square root of that to get that the distance between the 2 planes after 4 hours is

2745 miles

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vovangra [49]

Answer:

See below and the attachment

Step-by-step explanation:

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4 0

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

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,

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