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Lady bird [3.3K]

3 years ago

6

HELLPP!! Find the value of x that makes line m parallel to line n:

1 answer:

Veseljchak [2.6K]3 years ago

7 0

Answer:

x = 6

Step-by-step explanation:

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three consecutive even integers in which 20 more than twice of the third is 28 more than the sum of the first two integers

Doss [256]

Answer:

what is the answer choices?

5 0

2 years ago

Solve.<br> 92.5÷105 brainliest

Karo-lina-s [1.5K]

Answer:

$\frac{37}{42}$

Step-by-step explanation:

Given :

$\frac{92.5}{105}$

Multiply numerator and denominator by 2 :

$\frac{92.5}{105} \times\frac{2}{2}$

$\frac{185}{210}$ [Divide both sides by 5 as it is a common factor]

$\frac{185}{210} \div \frac{5}{5}$

$\frac{37}{42}$

6 0

2 years ago

Read 2 more answers

Which statements are true?

Rina8888 [55]

Answer:

The first one and the last one are true (Since −6 is 6 units to the left of 0, |−6|=6 and −6 is closer to zero on the number line than −7, so |−6|<|−7|.)

Step-by-step explanation:

Absolute value is the distance to zero, and it is always positive. That means the positive and negative versions of a number have the same absolute value. That means the first one is true, and the second and third ones are false. For the last one, -6 is closer to zero, so that means it would be true (The absolute value of -6 is 6, and the absolute value of -7 is 7).

4 0

3 years ago

Pls help i will give a good amount of points

AleksandrR [38]

The area is 10,994.39

3 0

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

3 years ago