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Nat2105 [25]
2 years ago
15

A line passes through the point (-2, 9) and has a slope of -3. Write an equation using the point-slope form for this. No Links.

Thank you
Mathematics
2 answers:
katrin [286]2 years ago
7 0

Answer:

(y-9)=-3(x+2)

Step-by-step explanation:

yeah-ya..... right?

jok3333 [9.3K]2 years ago
7 0

Answer:

y - 9 = -3(x + 2)

Step-by-step explanation:

Point-slope formula is named for the information that you can pick right out of the equation, that is a point on the line and the slope. It is:

y - y = m (x - x)

The m is the slope. They gave you that the slope is -3 so fill in -3 in place of the m.

The point you were given was (-2,9) the -2 is the x and the 9 is the y.

In the formula fill in the second y with 9 and fill in the second x with the -2.

Keep the first y and the first x (that's right inside the parenthesis) those stay as variables in your point-slope formula

It's literally a fill-in-the-blank problem.

y - 9 = -3(x + 2)

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9-3 divided by 1/3 +1
Andrei [34K]
Nine thirds? If so its would be 9/3 / 1/3?

The answer would be 1/3 / 1/3+1 making it 2/3

Im not sure so dont take my word
3 0
3 years ago
Read 2 more answers
What is the result of gaining 10 pound then losing 20
kiruha [24]

Answer:

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Step-by-step explanation:

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3 0
3 years ago
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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marshall27 [118]
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Answer:

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