No. Of roots in eq 8sec^x-6secx+1=0

1 answer:

7 0

Ok, let's assume it's "sec^2 x"

$8\sec^2x-6\sec x+1=0\\
 8\sec^2x-2\sec x-4\sec x+1=0\\
2\sec x(4\sec x-1)-1(4\sec x-1)=0\\
(2\sec x-1)(4\sec x-1)=0\\
2\sec x-1=0\\
2\sec x=1\\
\sec x=\dfrac{1}{2}\\
x\in\emptyset\\\\
4\sec -1=0\\
4\sec =1\\
\sec x =\dfrac{1}{4}\\
x\in\emptyset
$

So, the number of solutions (real ones) is 0.

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The quantity, Q, of a drug in the blood stream begins with 250 mg and decays to one-fifth its value over every 90 minute period.

Mazyrski [523]

Answer:

$a=250 \, mg\\\\b=5\\\\T= 90'$

Step-by-step explanation:

We have that $Q(t) = a\cdot b^{-\frac{t}{T}} \\$

Where t is the time (in minutes) and for the sake of dimensional consistency, let's assume that T is also in minutes, b is an adimensional number, and a is in mg. So we will have Q in mg as a consequence.

We now want to find out what values these constants might take. Let's see what happens when $t=0$ , that is, just as we start. At that point, we have that the amount of drug in the bloodstream must be equal to 250mg, thus:

$Q(0)= a\cdot b ^{-\frac{0}{T} }=250\,mg\\Q(0)= a=250\,mg$

We have found the constant $a$ ! It is the initial amount of drug! we have made use of the fact that any number raised to the 0th power is equal to one.

Now, we know that every 90 minutes, the amount of drug decreases to one fifth of its former value. How do we put this in mathematical form? Like so:

$Q(t+90')=Q(t)/5$

That is, 90 minutes after time t the amount of drug will be one fifth of the amount of drug at time t. Let's expand the last equation:

$Q(t+90')=Q(t)/5\\\\a\cdot b^{-\frac{t+90'}{T} }=a\cdot b^{-\frac{t}{T} }/5\\\\ b^{-\frac{t+90'}{T} }=b^{-\frac{t}{T} }/5\\\\b^{-\frac{t}{T} }b^{-\frac{90'}{T} }=b^{-\frac{t}{T} }/5\\\\b^{-\frac{90'}{T} }=\frac{1}{5}$

Now the last expression isn't enough to determine both T and b, but that also means that we have some freedom in how we choose them. What seems most simple is to pick $T=90'$ and thus we will get: