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Oksana_A [137]
3 years ago
14

Its in the link belowwwww its 8th grade math.. tysvm

Mathematics
1 answer:
Alexandra [31]3 years ago
5 0

Answer:

9

Step-by-step explanation:

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Determine the equation of the line that passes through (2,4) with a slope of -2/3
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Answer:

y = -2/3x + 16/3

Step-by-step explanation:

y = -2/3x + b

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4 = -4/3 + b

16/3 = b

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A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,
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We have been given that a huge ice glacier in the Himalayas initially covered an area of 45 square kilo-meters. The relationship between A, the area of the glacier in square kilo-meters, and t, the number of years the glacier has been melting, is modeled by the equation A=45e^{-0.05t}.

To find the time it will take for the area of the glacier to decrease to 15 square kilo-meters, we will equate A=15 and solve for t as:

15=45e^{-0.05t}

\frac{15}{45}=\frac{45e^{-0.05t}}{45}

\frac{1}{3}=e^{-0.05t}

Now we will switch sides:

e^{-0.05t}=\frac{1}{3}

Let us take natural log on both sides of equation.

\text{ln}(e^{-0.05t})=\text{ln}(\frac{1}{3})

Using natural log property \text{ln}(a^b)=b\cdot \text{ln}(a), we will get:

-0.05t\cdot \text{ln}(e)=\text{ln}(\frac{1}{3})

-0.05t\cdot (1)=\text{ln}(\frac{1}{3})

-0.05t=\text{ln}(\frac{1}{3})

t=\frac{\text{ln}(\frac{1}{3})}{-0.05}

t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-0.05\cdot 100}

t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-5}

t=-\text{ln}(\frac{1}{3})\cdot 20

t=-(\text{ln}(1)-\text{ln}(3))\cdot 20

t=-(0-\text{ln}(3))\cdot 20

t=20\text{ln}(3)

Therefore, it will take 20\text{ln}(3) years for area of the glacier to decrease to 15 square kilo-meters.

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3 years ago
Peter and John went shopping and spent ratio 2:3 If Peter spent $500 how much did John spend?
sveta [45]
John spent a total of $750
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3 years ago
Get 14 points, Plz help me with this question, and give the right answer cause it's important
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All angles are of triangle and thus give sum of 180⁰

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