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Amanda [17]
3 years ago
15

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

Mathematics
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.

Hope I helped!

Please mark Brainliest!

7 0
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

6 0
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

5 0
3 years ago
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Find each percent of change. Round to the nearest whole percent if necessary. State whether the percent of change is an increase
Alex787 [66]
So first we see if increase or decrease
180<210
it is increase

then we assume that
percent change
find how much change took place
210-180=30
30/180=x% change
percent means parts out ofo 100s so x%=x/100
30/180=x/100
multiply both sides by 100
3000/180=x
16.66666=x
answe ris 16.6%
7 0
3 years ago
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