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3 years ago

Simplify. 2 sixth, 2 ninth, 2 tenths

Mathematics

2 answers:

Viefleur [7K]3 years ago

8 0

Answer:

1/3, 2/9, 1/5

Step-by-step explanation:

2 sixths simplifies to 1/3, 2/9 cannot be simplified, and 2 tenths simplifys to 1/5

Vsevolod [243]3 years ago

5 0

Answer and Step-by-step explanation:

$\frac{2}{6} -> \frac{1}{3}\\\\\\\frac{2}{9} \\\\\\\frac{2}{10} ->\frac{1}{5}$

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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

Which number line shows the graph of -275/ -2 1 - 2 -- 2

Ray Of Light [21]

Answer: B

Step-by-step explanation: Since you know the number is not apart of the solution, the circle will definitely be open. But the circle is open so, it's apart of the problem

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3 years ago

A rectangle with a perimeter of 72 inches

attashe74 [19]

Answer:

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Step-by-step explanation:

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3 years ago

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Simplify your perimeter expression as much as possible.

Marysya12 [62]

Answer:

1019191991

Step-by-step explanation:

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2 years ago

Using the right triangle below find the cotangent of angle A

pogonyaev

Answer:

option d is correct

Step-by-step explanation:

cot=b/p

p=2

b=2 $\sqrt{3}$

so b/p=2 $\sqrt{3}$ /2= $\sqrt{3}$

7 0

4 years ago