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uranmaximum [27]
3 years ago
8

ALGEBRA 1 ! Please help!! need asap 100 POINTS

Mathematics
2 answers:
aleksley [76]3 years ago
5 0

Answer:

1. Skyler's processes are incorrect. She put (3^6)^x = 9, but when working with exponential problems, both sides have to have the same base. 3 does not equal nine, so you can automatically rule out her approach.

2. Robert's processes are correct. He set both bases equal to each other and then solved for the exponents.

3. Kaitlyn is also correct, she set the bases equal to 3 and solved for the exponents.

Step-by-step explanation:

When working with exponents, the goal is to get the bases equal to each other. In this case, you could have set it equal to 9 (which Robert did) or you could have set it equal to 3 (like Kaitlyn did). But you never want the bases to be conflicting, which is what Skyler did. Just by looking at the problem and using deductive reasoning, it is likely that Robert and Kaitlyn are both correct, since they have the same answers, although they used different methods, but it is still important to look at the steps.

Skyler's one mistake was setting 3 = 9, because that is impossible. If you plug in her answer of 3/2 into x, you get a ridiculous number of 19,683 = 9, which is not possibly true. For the others, when you plug in 1/3, you get 729^(1/3) = 9, so 9=9.

Hopefully this makes sense, if not, I can try to explain it better. Thanks much & have a wonderful day.

zhannawk [14.2K]3 years ago
4 0

Answer:

1. Skyler's processes are incorrect. She put (3^6)^x = 9, but when working with exponential problems, both sides have to have the same base. 3 does not equal nine, so you can automatically rule out her approach.

2. Robert's processes are correct. He set both bases equal to each other and then solved for the exponents.

3. Kaitlyn is also correct, she set the bases equal to 3 and solved for the exponents.

Step-by-step explanation:

When working with exponents, the goal is to get the bases equal to each other. In this case, you could have set it equal to 9 (which Robert did) or you could have set it equal to 3 (like Kaitlyn did). But you never want the bases to be conflicting, which is what Skyler did. Just by looking at the problem and using deductive reasoning, it is likely that Robert and Kaitlyn are both correct, since they have the same answers, although they used different methods, but it is still important to look at the steps.

Skyler's one mistake was setting 3 = 9, because that is impossible. If you plug in her answer of 3/2 into x, you get a ridiculous number of 19,683 = 9, which is not possibly true. For the others, when you plug in 1/3, you get 729^(1/3) = 9, so 9=9.

Hopefully this makes sense, if not, I can try to explain it better. Thanks much & have a wonderful day.

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You have 8,640 grams of a radioactive kind of cesium. If its half-life is 30 years, how much
Vikki [24]

The mass of cesium left afer 120 years if the original mass of cesium is 8640 with an half-life of 30 years,  is 540 grams.

<h3>What is half-life?</h3>

This can be defined as the time taken for half the amount of a radioactive substance to decay.

To calculate the amount of cesium left after 120 years, we use the formua below.

Formula:

  • R' = R/(2^{T/t})............. Equation 1

Where:

  • R = Original amount of Cesium
  • R' = New amount of cesium
  • T = Total time
  • t = Half-life

From the question,

Given:

  • R = 8640
  • T = 120 years
  • t = 30 years

Substitute these values into equation 1

  • R' = 8640/(2^{120/30})
  • R' = 8640/2^{4}
  • R' = 8640/16
  • R' = 540 grams.

Hence, The mass of cesium left afer 120 years is 540 grams.

Learn more about half-life here: brainly.com/question/25750315

4 0
2 years ago
The half-life is a substance is 375 years. If 70 grams is present now, how much will be present in 500 years?
Lynna [10]

Answer:

27.76 grams will be present in 500 years

Step-by-step explanation:

The given formula is A=A_{o}e^{kt} , where A is the value of the substance in t years, and A_{o} is the initial value

∵ The half-life is a substance is 375 years

- Substitute A by \frac{1}{2}A_{o} and t by 375 to find the value of k

∴ \frac{1}{2}A_{o}=A_{o}e^{375k}

- Divide both sides by A_{o}

∴ \frac{1}{2}=e^{375k}

- Insert ㏑ in both sides

∴ ㏑( \frac{1}{2} ) = ㏑ ( e^{375k} )

- Remember ㏑ ( e^{n} ) = n

∵ ㏑ ( e^{375k} ) = 375 k

∴ ㏑( \frac{1}{2} ) = 375 k

- Divide both sides by 375

∴ k ≈ -0.00185

∴  A=A_{o}e^{-0.00185t}

∵ 70 grams is present now

- That means the initial value is 70 grams

∴ A_{o} = 70

∵ The time is 500 years

∴ t = 500

- Substitute the values of A_{o} and t in the formula

∵ A=70e^{-0.00185(500)}

∴ A = 27.76

∴ 27.76 grams will be present in 500 years

3 0
3 years ago
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