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luda_lava [24]
3 years ago
7

Need help please fast and urgent

Mathematics
1 answer:
Airida [17]3 years ago
6 0
The answer is b bro i belive

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What is the solution to the system of equations?
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Answer:

(2, - 4 )

Step-by-step explanation:

Given the 2 equations

x + 2y = - 6 → (1)

3x + y = 2 → (2)

Multiplying (2) by - 2 and adding to (1) will eliminate the term in y

- 6x - 2y = - 4 → (3)

Add (1) and (3) term by term to eliminate y

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x = 2

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8 0
4 years ago
A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,
guajiro [1.7K]

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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