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Bumek [7]
3 years ago
7

Slop of (-1,3) and (2,-3)

Mathematics
2 answers:
anyanavicka [17]3 years ago
5 0

Answer: -2

Step-by-step explanation:

The equation for finding the slope is the change in y divided by the change in x, so you plot the numbers

(y2-y1/x2-x1) -> (-3-3/2-(-1)) -> (-6/3) -> -2

And so the slope is -2

If you have any more questions on finding the slope, you can ask! :)

const2013 [10]3 years ago
4 0

Answer: -2

Step-by-step explanation:

slope= y2-y1 / x2-x1

m= -3-3 / 2-(-1)

m= -6 / 3

m= -2

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bekas [8.4K]
Gonna do a little subbing here...

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40(3) - 0.05(40(3)) = T
120 - 0.05(120) = T.....120 - 6 = T.....114 = T
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for 40 pgs of basic proof reading....c = 3.95
40(3.95) - 0.05(40(3.95) = T
158 - 0.05(158) = T....158 - 7.9 = T.....150.10
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for 40 pgs of extended proof reading....c = 5
40(5) - 0.05(40(5) = T
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3 0
3 years ago
Find the interquartile range of the data. 68, 15, 55, 5, 66, 42, 51, 12, 23 *
Elan Coil [88]

Answer:

72.5

Step-by-step explanation:

5, 12, 15, 23, 42, 51, 55, 66, 68

42 = median

Q1 median = 21

Q2 median = 93.5

93.5 - 21 = 72.5

         Hope this helped!!!

4 0
3 years ago
Find the<br> area of the figure.
EastWind [94]

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Step-by-step explanation:

8 0
3 years ago
3. Antony and Cleopatra are 100 miles apart . To meet at a planned location , Antony travels at 12 mph and Cleopatra travels at
Vlad1618 [11]

Answer:

60 miles

Step-by-step explanation:

Distance apart = 100

Rate of travel. :

Anthony = 12 mph

Cleopafda = 8 mph

Using the relation :

Speed = distance / time

Distance = speed * time

If they leave at the same time, travel time Can be represented as x

Anthony's distance + Cleopafda distance = total distance

12x +. 8x = 100

20x = 100

x = 5

Hence, they both traveled for 5 hours before meeting.

Distance covered by Anthony :

Speed * time

12 mph * 5h = 60 miles

Anthony must travel. For 60 miles.

5 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} &#10;&#10;2 =   e^{ \alpha t} &#10;&#10; \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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