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Colt1911 [192]
3 years ago
8

I need help with this too why is math so hard ?

Mathematics
1 answer:
ira [324]3 years ago
6 0
Question 7;
20 mg iodine is given. Since the half life of it 8 days,
It will reduce to 10mg after 8 days. ( 20/2 =10mg)
It will reduce to 5mg after another 8 days. (10/2 =5mg)
It will reduce to 2.5mg after another 8 days.(5/2 =2.5mg)
And it will reduce to 1.25mg after another 8 days.

So after 32 days which is 4 * 8 days, it will be only 1.25mg left.

Question 8a ;
Growth factor is 3 because you multiply every number to get next one. 3*3 =9 , 9*3=27 , 27*3=81
Growth rate is %300 because it is 3 times of the actual number. %100 of number 3 is 3, %200 of number 3 is 6 and %300 of number 3 is 9.

Question 8b;
Model is the first one, y=3^x
Because sequence is orders of 3. 3 order of 1 is 3, 3 order of 2 is 9 , 3 order of 3 is 27.
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Andreyy89

Answer:

\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx}= x - \ln(1 + e^{x}) + C\end{aligned}.

Step-by-step explanation:

The first derivative of the denominator 1 + e^{x} is e^{x}. Rewrite the fraction to obtain that expression on the numerator.

\begin{aligned}\frac{1}{1 + e^{x}} &= \frac{1 + e^{x}}{1 + e^{x}} - \frac{e^{x}}{1+e^{x}}\\&=1-\frac{e^{x}}{1+e^{x}}\end{aligned}.

In other words,

\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx} &= \int{dx} - \int{\frac{e^{x}}{1+e^{x}}\cdot dx}\end{aligned}.

Apply u-substitution on the integral \displaystyle \int{\frac{e^{x}}{1+e^{x}}\cdot dx}:

Let u = 1 + e^{x}. u > 1.

du = e^{x}\cdot dx.

\displaystyle \int{\frac{e^{x}}{1+e^{x}}\cdot dx} = \int{\frac{du}{u}} = \ln{|u|} = \ln{u} +C = \ln{(1+e^{x})}+C.

Therefore

\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx} &= \int{dx} - \int{\frac{e^{x}}{1+e^{x}}\cdot dx}\\ & = x - \ln{(1 + e^{x})}+C\end{aligned}.

7 0
3 years ago
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