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3 years ago

15

Racionalize o denominador da equação abaixo: 5 + 3 ²√5 / ²√5

1 answer:

GrogVix [38]3 years ago

4 0

I hope this helps

$\cfrac{5+3 \sqrt{5} }{ \sqrt{5} } =\cfrac{(5+3 \sqrt{5}) \sqrt{5} }{ \sqrt{5}* \sqrt{5} }= \cfrac{5 \sqrt{5}+15 }{5} = \cfrac{5 (\sqrt{5}+3) }{5} = \sqrt{5}+3$

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What number would go in the blank? (3 • 5) • 2 = 3 • (__ • 2)<br><br> 10<br><br> 5<br><br> 2

tester [92]

Answer:

5

Step-by-step explanation:

It's the one that's missing.

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2 years ago

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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

The width A rectangle is 3 cm less then its length. The perimeter of the rectangle is 30 CM. Find The length and width of the re

Tema [17]

w = l - 3

w = 9 - 3

w = 6

//

l = 9

w = 6

//

p = 2(9) + 2(6)

p = 30

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2 years ago

Find two consecutive odd integers whose sum is 96

jekas [21]

Answer: 31 and 33

Step-by-step explanation: THIS IS NOT A VERIFIED ANSWER SO IF ITS WRONG IM AM TERIBLY SORRY BUT CAN I PLEASE GET BRAINLIEST

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Alex787 [66]

So first we see if increase or decrease
180<210
it is increase

then we assume that
percent change
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210-180=30
30/180=x% change
percent means parts out ofo 100s so x%=x/100
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multiply both sides by 100
3000/180=x
16.66666=x
answe ris 16.6%

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3 years ago