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AlladinOne [14]
3 years ago
10

Function 1 is represented by the equation y=-4/5x-2,and function 2 is represented by the graph above

Mathematics
1 answer:
AlexFokin [52]3 years ago
6 0
B) only fuction 1 is the answer.
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n-the\ number\\\\the\ quotient\ of\ a\ number\ (n)\ and\ 12=\boxed{n:12=\frac{n}{12}}
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Estimate each Quotient
Doss [256]
Okay, through simple division I'll give you the long side of the work. Known as Long Division: [This may turn out unaligned correctly, but hopefully it works.]

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73 / 59628
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Question 2:
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Question 3:
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I hope that helps, have a great rest of your day! ^ ^
{-Ghostgate-}
3 0
3 years ago
I need help with the area of this figure
Ahat [919]

Answer:

31.63

Step-by-step explanation:

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3 0
3 years ago
At the beginning of an experiment, a scientist has 300 grams of radioactive goo. After 150 minutes, her sample has decayed to 37
monitta

Answer:

Half-life of the goo is 49.5 minutes

G(t)= 300e^{-0.014t}

191.7 grams of goo will remain after 32 minutes

Step-by-step explanation:

Let M_0\,,\,M_f denotes initial and final mass.

M_0=300\,\,grams\,,\,M_f=37.5\,\,grams

According to exponential decay,

\ln \left ( \frac{M_f}{M_0} \right )=-kt

Here, t denotes time and k denotes decay constant.

\ln \left ( \frac{M_f}{M_0} \right )=-kt\\\ln \left ( \frac{37.5}{300} \right )=-k(150)\\-2.079=-k(150)\\k=\frac{2.079}{150}=0.014

So, half-life of the goo in minutes is calculated as follows:

\ln \left ( \frac{50}{100} \right )=-kt\\\ln \left ( \frac{50}{100} \right )=-(0.014)t\\t=\frac{-0.693}{-0.014}=49.5\,\,minutes

Half-life of the goo is 49.5 minutes

\ln \left ( \frac{M_f}{M_0} \right )=-kt\Rightarrow M_f=M_0e^{-kt}

So,

G(t)= M_f=M_0e^{-kt}

Put M_0=300\,\,grams\,,\,k=0.014

G(t)= 300e^{-0.014t}

Put t = 32 minutes

G(32)= 300e^{-0.014(32)}=300e^{-0.448}=191.7\,\,grams

7 0
3 years ago
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