Ask question

Ask question

All categories

3 years ago

10

How to Simplify the rational expression.

1 answer:

alexandr402 [8]3 years ago

3 0

2r - 4 2(r - 2)
--------- = ----------- = 2
r - 2 r - 2

answer
2

You might be interested in

Merl is a Canadian exchange student in China, and he currently has 12,000 yuan in his bank account. If the exchange rate changes

V125BC [204]

The value in Canadian dollars of the money in Merl's bank account goes down because it takes .14 more Yuan to make 1 Canadian dollar.

3 0

2 years ago

‼️‼️‼️‼️‼️‼️‼️‼️‼️‼️‼️‼️PLEASE PLEASE HELP I HAVE BEEN WAITING FOR OVER AN HOUR

Gnoma [55]

I cant see the screen clearly

6 0

2 years ago

PLEASE HELP Which of the following is false about the Y intercept

Flauer [41]

The answer choice that is false is D.

The form of the y-intercept is (0,Y). Remember that x comes first in an ordered pair and y comes second.

Hope this helps!

7 0

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

What is the sum of (1.3 × 10-7) and (-3.8 × 10-3)?

SVETLANKA909090 [29]

b. -2.5 × 10-4 is the answer

3 0

2 years ago