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3 years ago

Solve each equation 2/3 (m - 6)=3

2 answers:

8 0

Hey there! :D

2/3(m-6)= 3

2/3m-4=3

2/3m= 7

m= 10.5

We can check it.

2/3(10.5-6)= 3

7-4=3

3=3

I hope this helps!
~kaikers

Vitek1552 [10]3 years ago

7 0

2/3(m-6)=3

multiply 2 by everything in fraction; keep same denominator for everything in parentheses
(2*m)/3 - (2*6)/3= 3
2m/3 - 12/3= 3

simplify 12/3= 4
2m/3 - 4= 3

Multiply everything by 3 to eliminate fraction
(3)(2m/3) - (3)(4)= (3)(3)
2m - 12= 9

add 12 to both sides
2m= 21

divide both sides by 2
m= 10.5

Check:
2/3(m-6)=3
2/3(10.5-6)= 3
2/3(4.5)= 3
(2*4.5)/3= 3
9/3= 3
3= 3

Hope this helps! :)

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4 years ago

Rutherfordium- 265 is a radioactive substance that has a half-life of 13 hours. If there is a 149g sample of Rutherfordium- 265

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Answer:

3.21 grams

Step-by-step explanation:

First we are finding the radioactive decay constant using the formula:

lambda = $\frac{ln(2)}{half-life}$

where

lambda is the radioactive decay constant.

$half-life$ is the half life of the radioactive substance.

We know from our problem that Rutherfordium- 265 has a half-life of 13 hours, so let's replace the value in our formula.

lambda = $\frac{ln(2)}{13hours}$

lambda = 0.0533 per hour

Now we can use the decay formula to find the remaining quantity of the substance:

$N_t=N_0e^{(-lamda)(t)$

where

$N_t$ is the ending amount

$N_0$ is the beginning amount

$t$ is the time (in hours)

We know from our problem that there is a 149g sample of Rutherfordium- 265 left at 12am on October 19th, so $N_0=149$ . Notice that there are exactly 3 days from 12 am October 19th to 12 am October 22nd, so we have $3days(\frac{24hours}{1day} )=72hours$ ; therefore $t=72$ . Now we can replace all the values in our formula: