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stepan [7]
3 years ago
8

How do i solve 4(x+8) - 4= 34 - 2x ?

Mathematics
2 answers:
mylen [45]3 years ago
7 0
If you have any questions about how I did this just message me.

Hope this helps :)

hoa [83]3 years ago
5 0
X= -157 to get that first you need to get 4×88 - 4 = 34 -2x then you do 352 - 4 = 34 - 2x after that do 348 = 34 - 2x then right 2X = 34 - 348 then the last step you do is 2x = -134
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2 years ago
Read 2 more answers
The concentration C of certain drug in a patient's bloodstream t hours after injection is given by
frozen [14]

Answer:

a) The horizontal asymptote of C(t) is c = 0.

b) When t increases, both the numerator and denominator increases, but given that the grade of the polynomial of the denominator is greater than the grade of the polynomial of the numerator, then the concentration of the drug converges to zero when time diverges to the infinity. There is a monotonous decrease behavior.  

c) The time at which the concentration is highest is approximately 1.291 hours after injection.

Step-by-step explanation:

a) The horizontal asymptote of C(t) is the horizontal line, to which the function converges when t diverges to the infinity. That is:

c = \lim _{t\to +\infty} \frac{t}{3\cdot t^{2}+5} (1)

c = \lim_{t\to +\infty}\left(\frac{t}{3\cdot t^{2}+5} \right)\cdot \left(\frac{t^{2}}{t^{2}} \right)

c = \lim_{t\to +\infty}\frac{\frac{t}{t^{2}} }{\frac{3\cdot t^{2}+5}{t^{2}} }

c = \lim_{t\to +\infty} \frac{\frac{1}{t} }{3+\frac{5}{t^{2}} }

c = \frac{\lim_{t\to +\infty}\frac{1}{t} }{\lim_{t\to +\infty}3+\lim_{t\to +\infty}\frac{5}{t^{2}} }

c = \frac{0}{3+0}

c = 0

The horizontal asymptote of C(t) is c = 0.

b) When t increases, both the numerator and denominator increases, but given that the grade of the polynomial of the denominator is greater than the grade of the polynomial of the numerator, then the concentration of the drug converges to zero when time diverges to the infinity. There is a monotonous decrease behavior.  

c) From Calculus we understand that maximum concentration can be found by means of the First and Second Derivative Tests.

First Derivative Test

The first derivative of the function is:

C'(t) = \frac{(3\cdot t^{2}+5)-t\cdot (6\cdot t)}{(3\cdot t^{2}+5)^{2}}

C'(t) = \frac{1}{3\cdot t^{2}+5}-\frac{6\cdot t^{2}}{(3\cdot t^{2}+5)^{2}}

C'(t) = \frac{1}{3\cdot t^{2}+5}\cdot \left(1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} \right)

Now we equalize the expression to zero:

\frac{1}{3\cdot t^{2}+5}\cdot \left(1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} \right) = 0

1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} = 0

\frac{3\cdot t^{2}+5-6\cdot t^{2}}{3\cdot t^{2}+5} = 0

5-3\cdot t^{2} = 0

t = \sqrt{\frac{5}{3} }\,h

t \approx 1.291\,h

The critical point occurs approximately at 1.291 hours after injection.

Second Derivative Test

The second derivative of the function is:

C''(t) = -\frac{6\cdot t}{(3\cdot t^{2}+5)^{2}}-\frac{(12\cdot t)\cdot (3\cdot t^{2}+5)^{2}-2\cdot (3\cdot t^{2}+5)\cdot (6\cdot t)\cdot (6\cdot t^{2})}{(3\cdot t^{2}+5)^{4}}

C''(t) = -\frac{6\cdot t}{(3\cdot t^{2}+5)^{2}}- \frac{12\cdot t}{(3\cdot t^{2}+5)^{2}}+\frac{72\cdot t^{3}}{(3\cdot t^{2}+5)^{3}}

C''(t) = -\frac{18\cdot t}{(3\cdot t^{2}+5)^{2}}+\frac{72\cdot t^{3}}{(3\cdot t^{2}+5)^{3}}

If we know that t \approx 1.291\,h, then the value of the second derivative is:

C''(1.291\,h) = -0.077

Which means that the critical point is an absolute maximum.

The time at which the concentration is highest is approximately 1.291 hours after injection.

5 0
2 years ago
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