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GarryVolchara [31]

3 years ago

12

Please Help..............................................................

1 answer:

maw [93]3 years ago

5 0

The second, 4th and 6th ones are the correct answers.

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Find the value of x in the figure

MA_775_DIABLO [31]

Answer:

x = 100

Step-by-step explanation:

$57 \degree + (x + 23) \degree = 180 \degree..(straight \: line \: \angle s) \\ (x + 80) \degree = 180 \degree \\ x + 80 = 180 \\ x = 180 - 80 \\ \huge \red{ \boxed{x = 100}}$

6 0

4 years ago

Read 2 more answers

Josie wants to make a drawstring backpack for her friends. She has 6 friends and the total for the backpacks is 48 dollars

Aneli [31]

8 dollars for each friend

7 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

4 years ago

Suppose a wild fire is doubling in size every 6 hours. If the initial size of the fire was 1 acre, how many acres will the fire

maria [59]

Step-by-step explanation:

2 days= 24 hours in 1 day, so multiply that by 2 to get 48 hours total.

Now find the amount of time the wild fire doubles by dividing 48 total hours by 6. So it will double 8 times

The size was 1 acre. Let's multiply by 2, 8 times

1

2

4

8

16

32

64

128

The fire will cover 128 acres in two days.

8 0

3 years ago

How does the function f(x) = -(x-4)^3 + 2 compare to the parent function f(x) = x^3? Please show and explain all your steps. Tha

Fynjy0 [20]

Translate by vector (4, -2) and reflect over x axis

translate by vector (4, -2) and reflect over x axis
translate by (4, -2) gives (x-4)^2 - 2
reflect over x axis means * -1 = -(x-4)^3 + 2

Mark brainliest please

4 0

3 years ago