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3 years ago

5

How many inches are in a meter.!!!!!!!!!!!!!!!!!!!??????????? I will mark brainliest.

2 answers:

7nadin3 [17]3 years ago

7 0

Answer:

39.3701

Step-by-step explanation:

Hope this helps!

~CoCo

Allushta [10]3 years ago

5 0

Answer:

in one meter there are <u><em>39.3701 inches </em></u>:)

Step-by-step explanation:

<em>please mark as brainliest tysm</em>

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Which autocad command can we use to quickly calculate the area and perimeter of a closed region defined by a polyline?

polet [3.4K]

The answer is: The Area command.

The explanation is shown below:

1. The Area command is a very useful command in AutoCad, and you can use it to calculate the area and the perimeter of a closed region draw with a polyline.

2. To use this command, you only need to type AREA and press the enter button. Then, you must select the points of the closed region and press enter again. Once you do this, the software will show you the perimeter and the area.

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4 years ago

a motorcycle shop maintains an inventory of four times as many new bikes as used bikes. if there are 64 new bikes,how many used

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3 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

4 years ago

2x-3y> or equal to 12

Drupady [299]

Answer:

it's not greater than it would be 2x-3y because you can add then together because the not the same variable

Step-by-step explanation:

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4 years ago

What is the solution set of the following quadratic equation?

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3 years ago