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anastassius [24]
3 years ago
9

Uma wrote the following. 15=x x=15 Which property of equality did she use?

Mathematics
2 answers:
Goryan [66]3 years ago
8 0

Answer:

The answer is - symmetric property of equality

Step-by-step explanation:

Uma wrote the following.

15=x; x=15

Transitive property of equality states that if a = b and b = c, then a = c. So, this is not the answer.

Substitution property of equality states that if x = y, then x can be substituted in for y and y can be substituted for x in any equation. This is not the answer.

Multiplication property of equality states that if you multiply both sides of an equation by the same number, the sides remain equal. So, this is not the answer.

Symmetric property of equality states that if a = b then b = a. So, this is the answer.

lbvjy [14]3 years ago
5 0
That would be the symmetric property of equality
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Step-by-step explanation:

Let the number of hours Nick does the job alone to be x - 3 hours

Hence, in 1 hour, Nick has done 1/x - 3

Together they complete the job in 9 hours

Hence, let's represent the job as 1

9/x + 9/x - 3 = 1

Multiply through by (x)(x-3)

9(x- 3) + 9(x) = 1

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Step-by-step explanation:

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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




4 0
3 years ago
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