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siniylev [52]
2 years ago
7

c{2}{4} + \frac{1}{3} " alt=" \frac{2}{4} + \frac{1}{3} " align="absmiddle" class="latex-formula">
I'm not sure how to do it​
Mathematics
2 answers:
Scilla [17]2 years ago
5 0
5/6 is the correct answer
Alchen [17]2 years ago
3 0

Answer:

\frac{5}{6}

Step-by-step explanation:

\frac{2}{4}+\frac{1}{3}\\Find a common denominator\\2*3=6\\4*3=12\\1*4=4\\3*4=12\\\frac{6}{12}+\frac{4}{12}\\6+4=10\\\frac{10}{12}=\frac{5}{6}

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Which equation represents the line that is parallel to y= 3/4x + 7 and passes through (-12,36)?
Nady [450]
In order for two lines to be parallel, the slope of the equations have to be the same.
y = 3/4x + 7 —> 3/4 is the slope.
For the new equation, the line has to go through (-12,36) which means that the y- intercept has to be 36.
The new equation would be y = 3/4x + 36.
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4 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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