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2 years ago

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I want to i want to i want to

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Rudik [331]2 years ago

7 0

Answer:

What do you mean?

Step-by-step explanation:

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The radius of a circle is 56 cm. Find the circumference of the circle to the nearest tenth. 351.9 cm 9,852.0 cm 112.0 cm 88.0 cm

borishaifa [10]

Alright, lets get started.

We have given a circle which has 56 cm radius.

We are asked to find its circumference.

The formula for circumference of circle = 2 pi r

Putting the value of pi and r

So the circumference = 2 * 2.142 * 56

Simplifing

Hence the circumference of circle = 351.9 cms

Answer is 351.9 cm

Hope it will help :)

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2 years ago

What is the value of 5t – 5? A. 0 B. 20 C. 45 D. 50

poizon [28]

Factor 5 de 5t - 5
5(t - 1)

¡Espero que esto ayude! :)

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2 years ago

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Subtract 7x + 2 from –5x2 – 2x.

Aleks [24]

Answer:

5x^2 + 9x + 2

Step-by-step explanation:

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2 years ago

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A family eats at a restaurant. The bill is $42. The family leaves a tip and spends $49.77.

jeka94

Answer;

7.77

Step-by-step explanation:

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2 years ago

The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nu

gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt} =-\lambda N$

$\Rightarrow \frac{dN}{N} =-\lambda dt$

Integrating both sides

$\int \frac{dN}{N} =\int\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$\Rightarrow ln|N_0|=-\lambda .0+c$

$\Rightarrow c= ln|N_0|$

$\therefore ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$ .......(1)

$\frac{N}{N_0}=e^{-\lambda t}$ .........(2)

Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
$ln|ab|=ln|a|+ln|b|$

$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
$ln|1|=0$

130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

$ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130$

$\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130$

$\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130$

$\rightarrow -ln|2|=-\lambda \times 130$

$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

We need to find the time when the sample remains 80% of its original.

$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

$\Rightarrow t\approx 42$

We can use the sample about 42 days.

7 0

3 years ago