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Iteru [2.4K]
3 years ago
12

The sum of two numbers is 72 and their difference is 26. find two numbers?

Mathematics
1 answer:
Archy [21]3 years ago
7 0

9514 1404 393

Answer:

  23, 49

Step-by-step explanation:

Let s represent the smaller of the two numbers. Then the larger is s+26, and the sum is ...

  s + (s+26) = 72

  2s = 72 -26

  s = (72 -26)/2 = 23

Then the larger number is

  s +26 = 49

The two numbers are 23 and 49.

_____

The reason we wrote the solution this way is to show you that the smaller number is half the difference between the given sum (72) and difference (26). This is the generic solution to a "sum and difference" problem. The smaller number is always half the difference of the give sum and difference. (Similarly, the larger is always half the sum of the given sum and difference.)

That is, the two numbers are always half the sum ± half the difference:

  72/2 ± 26/2 = {36+13, 36-13} = {49, 23}

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Since at t=0, n(t)=n0, and at t=∞, n(t)=0, there must be some time between zero and infinity at which exactly half of the origin
Airida [17]
Answer: t-half = ln(2) / λ ≈ 0.693 / λ

Explanation:

The question is incomplete, so I did some research and found the complete question in internet.

The complete question is:

Suppose a radioactive sample initially contains N0unstable nuclei. These nuclei will decay into stable nuclei, and as they do, the number of unstable nuclei that remain, N(t), will decrease with time. Although there is no way for us to predict exactly when any one nucleus will decay, we can write down an expression for the total number of unstable nuclei that remain after a time t:

N(t)=No e−λt,

where λ is known as the decay constant. Note that at t=0, N(t)=No, the original number of unstable nuclei. N(t) decreases exponentially with time, and as t approaches infinity, the number of unstable nuclei that remain approaches zero.

Part (A) Since at t=0, N(t)=No, and at t=∞, N(t)=0, there must be some time between zero and infinity at which exactly half of the original number of nuclei remain. Find an expression for this time, t half.

Express your answer in terms of N0 and/or λ.

Answer:

1) Equation given:

N(t)=N _{0} e^{-  \alpha  t} ← I used α instead of λ just for editing facility..

Where No is the initial number of nuclei.

2) Half of the initial number of nuclei: N (t-half) =  No / 2

So, replace in the given equation:

N_{t-half} =  N_{0} /2 =  N_{0}  e^{- \alpha t}

3) Solving for α (remember α is λ)

\frac{1}{2} =  e^{- \alpha t} 

2 =   e^{ \alpha t} 

 \alpha t = ln(2)

αt ≈ 0.693

⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ




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3 years ago
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Step-by-step explanation:

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2 years ago
The solution to a quadric equation are x=2 and x=7. If k is a nonzero constant which of the following must be equal to the quadr
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Answer:

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Step-by-step explanation:

Given in the question as ,

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