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3 years ago

5

Need help pls Find m

1 answer:

Gekata [30.6K]3 years ago

8 0

Answer:

m < its right here have fun on the test

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Use diagrams of equivalent ratios to show that the ratios 2:3,4:6, and 8:12 are equivalent

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Can someone please go over to my page and help me with this problem, its worth 99 points,no joke

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3 years ago

(3x - 2)³= <br>gives the solution pls.

Nonamiya [84]

Answer:

27x^3 - 54x^2 + 36x - 8

Step-by-step explanation:

(3x - 2)^3

= (3x - 2) (3x - 2) (3x - 2)

= 27x^3 - 54x^2 + 36x - 8

Hope this helps :)

Let me know if there are any mistakes!!

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3 years ago

Read 2 more answers

Bryan bought 10 daisies and 5 sunflowers. Each daisy cost $2 and each sunflower cost $5. Bryan had $12 left after he made his pu

sasho [114]

Answer:

Bryan spent $45 on flowers. The last sentence isn't needed to answer this question.

Step-by-step explanation:

If you multiply 10 by 2, then 5 by 5, and add it at the end you get 45.

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3 years ago

Marks test scores for this semester were 83,80,91,74,66,95,87,72,95,97 what is the mode of his test scores

Vlada [557]

Answer:

95

Step-by-step explanation:

Mode is what is the most frequent number that occurred; 95 occurred the most from the data.

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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