Use your calculator to find tan 117° 35' 20"

Mathematics

1 answer:

Zarrin [17]3 years ago

5 0

C. -1.913

Step-by-step explanation:

Because i knew the answer.

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The half-life of a certain tranquilizer in the bloodstream is 22 hours. How long will it take for the drug to decay to 90% of

nasty-shy [4]

Answer:

It will take approximately 3.34 hours for the drug to decay to 90% of the original dosage

Step-by-step explanation:

As suggested, we use the formula for exponential decay:

$A(t)=A_0\,e^{-k\,t}$

From the given information, the half life of the drug in blood id 22 hours, so that means that it takes that number of hours to go from the initial value $A_0$ , to a final value equal to $A_0/2$ . Using this information we can find the decay rate "k" by solving for this parameter in the formula, and using the natural log function to bring the exponent down:

$A(t)=A_0\,e^{-k\,t}\\\frac{A_0}{2} =A_0\,e^{-k\,*22}\\\frac{A_0}{A_0*2} =e^{-k\,*22}\\\frac{1}{2} =e^{-k\,*22}\\ln(\frac{1}{2})=-k\,*22\\ k=-\frac{ln(\frac{1}{2})}{22} \\k=0.0315$

Now we use this value for the decay rate "k" to calculate how long it would take to decay to 90% of the original dose;

$A(t)=A_0\,e^{-0.0315\,t}\\0.9*A_0} =A_0\,e^{-0.0315\,t}\\\frac{0.9*A_0}{A_0} =e^{-0.0315\,t}\\0.9 =e^{-0.0315\,t}\\ln(0.9)=-0.0315\,t\\ t=-\frac{ln(0.9)}{0.0315} \\t=3.3447\,hours$

7 0

3 years ago

Please, I need help ASAP!

enot [183]

Answer:

Step-by-step explanation:

You can eliminate B and D because it’s using $\frac{g}{f}$ , but the question is asking $\frac{f}{g}$ . So it’s either A or C. They are both written correctly, but out of the two, A is correct because just that x ≠ $\sqrt{7}$ part of it is what makes it true.

The reason being is that the only thing that won’t make $\frac{2x+1}{x^{2}-7 }$ true is when the denominator equals 0, because anything divided by 0 is undefined. So, when you plug $\sqrt{7}$ for x in the denominator you get: $\sqrt{7^{2} } -7$ . The $\sqrt{7^{2} }$ part just turns into 7, and 7 - 7 is 0. So you don’t want x to equal $\sqrt{7}$ .