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DochEvi [55]
2 years ago
10

Element X decays radioactively with a half life of 7 minutes. If there are 980 grams of Element X, how long, to the nearest tent

h of a minute, would it take the element to decay to 40 grams?
Mathematics
2 answers:
xeze [42]2 years ago
8 0

Answer:

32.3m

Step-by-step explanation:

Use equation A = Ao(0.5)^(t/t½)

Now input values: 40 = 980(0.5)^(t/7)

rearrange for t: t = [7ln(2/49)]/ / ln(0.5). Then from here, you can extract 32.3 as your answer

GrogVix [38]2 years ago
7 0

Answer:

t = 32.3 min

Step-by-step explanation:

This is an example of half-life exponential decay.

For situations like this we have a formula.

N = N_{0} * (\frac{1}{2})^n

N represents the end amount of parent element, N (naught) represents the starting amount of the parent element, and n being the amount half-lifes that have occurred.

Now n, can be broken down further into t/p, which means the time elapsed (t) divided by time it takes for one decay period (p), which essentially gives you the amount of half-lifes that have occurred. So you can represent the formula as:

N = N_{0} * (\frac{1}{2})^{t/p}

The question tells you you start off with 980 grams of Element x, so put that for N naught. N = 980g * (\frac{1}{2})^{t/p}. Then it tells you half-life is 7 minutes, which means the period it takes for one half-life to occur is 7, which is p. N = 980g * (\frac{1}{2})^{t/7min}. You are also told that you end up with 40 grams, so put that in for N. 40g = 980g * (\frac{1}{2})^{t/7min}. Now we solve for t, the time elapsed.

(This is the part where you use logs to solve, but if you are still rough on this, I'll walk you through it)

Start off by dividing on both sides.

\frac{40g}{980g} = (\frac{1}{2})^{t/7min} \\, \frac{2}{49} = (\frac{1}{2})^{t/7min}

Take the log of both sides.

log(\frac{2}{49}) = log( (\frac{1}{2})^{t/7min})

Now we gotta get t out of the exponent, we'll do this by using log laws. More specifically the exponent law for log laws which states. log(b^x) = x*log(b). If we apply that to our situation like so.. log((\frac{1}{2})^{t/7min})  = \frac{t}{7min} * log(\frac{1}{2})

Now our whole equation now looks like this:

log(\frac{2}{49}) =   \frac{t}{7min} *  log( \frac{1}{2})

keep simplifying,

\frac{log(\frac{2}{49}) }{log(\frac{1}{2}) } = \frac{t}{7min}

7min *\frac{log(\frac{2}{49}) }{log(\frac{1}{2}) } = t\\\\t = 32.3min \\

Now you have your answer!

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An animal shelter houses only cats and dogs, there are 25% more cats these dogs. If there are 40 cats, how many dogs are there,
jok3333 [9.3K]

Answer:

There are 32 dogs and 72 animals in total.  

Step-by-step explanation:

In order to find the amount of dogs, you need to set up an equation to solve for the missing variable.  Let 'c' be the amount of cats and 'd' be the amount of dogs.  We know that there are 40 cats, so c = 40.  We also know that there are 25% more cats than dogs, so we can form an equation to express this difference:  1.25d = c, indicating that the number of cats is equal to the total number of dogs, plus an additional 25%.  Since we know c = 40, we can set our second equation equal to 40:  1.25d = 40.  Using inverse operations, we divide both sides by 1.25 to get d = 32.  So, d + c, or 32 + 40 = 72 total animals.  

6 0
3 years ago
1) Cathy borrows $3280 at 0.3% simple interest per month. When Cathy pays the loan back 9 years later, what is the total amount
aksik [14]

Answer:

Step-by-step explanation:

we calculate how much 0.3% is from 3280

y=3280*0.3/100=9.84$

months in 9y

12*9=108

the interest of 9y

108y=1062.72$

the sum of interest + the money borrowed

1062.72+3280=4342.72$

6 0
2 years ago
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Given the following figure. Which of the following is congruent with the figure above
tigry1 [53]
F and E are congruent with the figure above
3 0
3 years ago
If f(x)=x^3 which if the following describes the graph of f(x-3)
Katena32 [7]
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5 0
3 years ago
An urn contains two blue balls (denoted B1 and B2) and three white balls (denoted W1, W2, and W3). One ball is drawn from the ur
tatuchka [14]

Answer:

(a)10 Outcomes

(b)\dfrac{2}{5}

Step-by-step explanation:

An urn contains two blue balls (denoted B_1 \:and\: B_2) and three white balls (denoted W_1, W_2 \:and\: W_3).

In the selection, a ball is picked and replaced.

The possible outcomes of the experiment are:

B_1B_1,B_1B_2,B_1W_1,B_1W_2,B_1W_3\\B_2B_1,B_2B_2,B_2W_1,B_2W_2,B_2W_3\\W_1B_1,W_1B_2,W_1W_1,W_1W_2,W_1W_3\\W_2B_1,W_2B_2,W_2W_1,W_2W_2,W_2W_3\\W_3B_1,W_3B_2,W_3W_1,W_3W_2,W_3W_3

(a)If the first ball drawn is blue. the outcomes are:

B_1B_1,B_1B_2,B_1W_1,B_1W_2,B_1W_3\\B_2B_1,B_2B_2,B_2W_1,B_2W_2,B_2W_3

There are 10 outcomes if the first ball drawn is blue.

Probability that the first ball drawn is blue

=\dfrac{10}{25}\\ =\dfrac{2}{5}

4 0
3 years ago
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