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Oliga [24]
3 years ago
5

How do you simplify numbers with exponents

Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
7 0
You write it out then you’ll get the exponent
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Fill in the blank. I NEED AN ANSWER FASTTT
Yuri [45]

Answer:

It is =

Step-by-step explanation:

Because (6 x 1 = 6 ) + 1  x 1/100 = 6.1  

5 0
2 years ago
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.

6 0
3 years ago
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Hmm try 192.5 I think that’s right
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The mode of the data set: {10,2,8,9,5,2,6} is 2.
The reason why is because it’s the most frequent number being use. It occurs twice as you can see in the data set.
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3 years ago
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Tim goes on a car journey. For the first 15 minutes his average speed is 40 mph. He then stops for 25 minutes. He then completes
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Answer:

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