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svlad2 [7]
3 years ago
12

Iodine-131, a radioactive substance that is effective in locating brain​ tumors, has a​ half-life of only 8 days. A hospital pur

chased 24 grams of the substance but had to wait seven days before it could be used. How much of the substance was left after seven ​days?
Mathematics
1 answer:
Aloiza [94]3 years ago
7 0

The mass of substance left after 7 days is 13.09 g

The mass of substance left, N is given by

N = N₀exp(-λt) where λ = decay constant and N₀ = initial mass of substance present = 24 g and t = time

Also, λ = 0.693/t' where t' = half-life of iodine = 8 days

So, λ = 0.693/t'

λ = 0.693/8

λ = 0.086625/day

Since the mass of substance left is N = N₀exp(-λt) and we require the mass of substance after t = 7 days,

N = N₀exp(-λt)

N = 24 gexp(-0.086625/day × 7 days)

N = 24 gexp(-0.606375)

N = 24 g × 0.5453

N = 13.09 g

So, the mass of substance left after 7 days is 13.09 g

Learn more about radioactive decay here:

brainly.com/question/23705307

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3 years ago
Simplify (5^0+4^-0•5)^2​
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Answer:

anything raised to the power of zero= 1

(1+1/4^½)²

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What is the measure of an angle if it is 260 degrees less than six times its own complement?
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40 degrees

Step-by-step explanation:

Complementary angles have a sum of 90.

y = 90 - x

The angle is 260 degrees less than six times its complement.

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3 years ago
Question : In the given figure , ∆ APB and ∆ AQC are equilateral triangles. Prove that PC = BQ.
lorasvet [3.4K]

Answer:

See Below.

Step-by-step explanation:

We are given that ΔAPB and ΔAQC are equilateral triangles.

And we want to prove that PC = BQ.

Since ΔAPB and ΔAQC are equilateral triangles, this means that:

PA\cong AB\cong BP\text{ and } QA\cong AC\cong CQ

Likewise:

\angle P\cong \angle PAB\cong \angle ABP\cong Q\cong \angle QAC\cong\angle ACQ

Since they all measure 60°.

Note that ∠PAC is the addition of the angles ∠PAB and ∠BAC. So:

m\angle PAC=m\angle PAB+m\angle BAC

Likewise:

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Since ∠QAC ≅ ∠PAB:

m\angle PAC=m\angle QAC+m\angle BAC

And by substitution:

m\angle PAC=m\angle QAB

Thus:

\angle PAC\cong \angle QAB

Then by SAS Congruence:

\Delta PAC\cong \Delta BAQ

And by CPCTC:

PC\cong BQ

5 0
2 years ago
Read 2 more answers
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