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Fiesta28 [93]
3 years ago
6

What is the slop in the graph?

Mathematics
2 answers:
likoan [24]3 years ago
7 0

Answer:

-2/5

Step-by-step explanation:

rise/run you count how much it rises and how much it runs

Juliette [100K]3 years ago
5 0

Answer:

-2/5

Step-by-step explanation:

rise over run

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6k ÷ 7 = -3<br> Solve for k
fomenos

Answer:

k = -3.5

Step-by-step explanation:

6k ÷ 7 = -3

Multiply 7 to both sides

6k = -21

Divide both sides by 6

6k/6 = -21/6

k = -3.5

To check if the answer is correct or not, you can always plug in -3.5 for k in the equation.

6(-3.5) ÷ 7 = -3

-21 ÷ 7 = -3

-3 = -3

-3 equals -3 so it is correct that k = -3.5

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Does anyone know how to solve this
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Convert: 76.8°f = ________ °c<br> a. 21.5<br> b. 25<br> c. 170.24<br> d. 24.89
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The answer to convert 76.8 fahrenheit to Celsius is D
 

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One-third of 2c subtracted frome 5 times nine
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-44.8 is the answer.
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3 years ago
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Use the exponential decay​ model, Upper A equals Upper A 0 e Superscript kt​, to solve the following. The​ half-life of a certai
Akimi4 [234]

Answer:

It will take 7 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of the substance after t years,

A=A_0 e^{kt}

Where,

A_0 = Initial amount of the substance,

If the half life of the substance is 19 years,

Then if t = 19, amount of the substance = \frac{A_0}{2},

i.e.

\frac{A_0}{2}=A_0 e^{19k}

\frac{1}{2} = e^{19k}

0.5 = e^{19k}

Taking ln both sides,

\ln(0.5) = \ln(e^{19k})

\ln(0.5) = 19k

\implies k = \frac{\ln(0.5)}{19}\approx -0.03648

Now, if the substance to decay to 78​% of its original​ amount,

Then A=78\% \text{ of }A_0 =\frac{78A_0}{100}=0.78 A_0

0.78 A_0=A_0 e^{-0.03648t}

0.78 = e^{-0.03648t}

Again taking ln both sides,

\ln(0.78) = -0.03648t

-0.24846=-0.03648t

\implies t = \frac{0.24846}{0.03648}=6.81085\approx 7

Hence, approximately the substance would be 78% of its initial value after 7 years.

5 0
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