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anonymous

  • one year ago

The sides of a square are 2 to the power of 4÷9 inches long. what is the area of the square?

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  1. Vocaloid
    • one year ago
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    side = 2^(4/9) area = side^2 = ?

  2. anonymous
    • one year ago
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    Wait what

  3. anonymous
    • one year ago
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    How can I solve this??? @vocaloid

  4. Vocaloid
    • one year ago
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    it gives you the length of one side, so to find the area, you just square the length (side)*(side) = 2^(4/9) * 2^(4/9) = ?

  5. anonymous
    • one year ago
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    Do I times it 4 times or just twice @vocaloid

  6. Vocaloid
    • one year ago
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    twice..

  7. anonymous
    • one year ago
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    @vocaloid

  8. anonymous
    • one year ago
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    I got 4 to the power of 16/81

  9. Vocaloid
    • one year ago
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    not quite, we have to use the exponent rule here keep the base and add the exponents together 2^(4/9) * 2^(4/9) = 2^(4/9 + 4/9) = ?

  10. anonymous
    • one year ago
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    Oh ok

  11. anonymous
    • one year ago
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    Now I got 2 to the power of 8/9

  12. anonymous
    • one year ago
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    Is that right? @vocaloid

  13. Vocaloid
    • one year ago
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    yes, sorry for the late reply

  14. anonymous
    • one year ago
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    Do you you think you can help me with another question? @vocaloid

  15. Vocaloid
    • one year ago
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    sure

  16. anonymous
    • one year ago
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    Given the function f(x) = 5^x , section A is from = x = 0 to x = 1 and the section B is from x = 2 to x = 3. Part A: Find The average rate of change of each section Part B: How many times greater is the average rate of change of section B then section A? Explain why one rate of change is greater than the other. @vocaloid

  17. Vocaloid
    • one year ago
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    part A) we want two values 1. the rate of change between x = 0 and x = 1 and 2. the rate of change between x = 2 and x = 3 so, let's just tackle the first part for now the rate of change between x = 0 and x = 1 can be found using the formula [f(1)-f(0)]/(1-0)

  18. anonymous
    • one year ago
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    I got one

  19. Vocaloid
    • one year ago
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    not quite, let's take it bit by bit f(1) = ?

  20. anonymous
    • one year ago
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    Is it one

  21. Vocaloid
    • one year ago
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    not quite f(x) = 5^x f(1) = ?

  22. anonymous
    • one year ago
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    5 to the power of one

  23. Vocaloid
    • one year ago
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    right now f(0) = ?

  24. anonymous
    • one year ago
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    5 to the power of zero

  25. Vocaloid
    • one year ago
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    right, so [f(1)-f(0)]/(1-0) = ?

  26. anonymous
    • one year ago
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    Is it 1/1

  27. Vocaloid
    • one year ago
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    no... remember, we just calculated f(1) and f(0) f(1) - f(0) = ?

  28. anonymous
    • one year ago
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    5^1 - 5^0

  29. Vocaloid
    • one year ago
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    right, so [f(1)-f(0)]/(1-0) = ?

  30. anonymous
    • one year ago
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    Do I have to solve the exponent

  31. Vocaloid
    • one year ago
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    yes

  32. anonymous
    • one year ago
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    5-1/ 1-0 = 4/1

  33. Vocaloid
    • one year ago
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    right, and 4/1 = ?

  34. anonymous
    • one year ago
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    4

  35. Vocaloid
    • one year ago
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    right, now let's calculate the rate of change from x = 2 to x = 3 [f(3)-f(2)]/(3-2) = ?

  36. anonymous
    • one year ago
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    5^3 - 5^2/ 3-2

  37. Vocaloid
    • one year ago
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    right, keep going...

  38. anonymous
    • one year ago
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    125 - 25/ 3-2

  39. Vocaloid
    • one year ago
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    right, keep going...

  40. anonymous
    • one year ago
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    100/1= 100

  41. Vocaloid
    • one year ago
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    right

  42. Vocaloid
    • one year ago
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    now we're finished w/ part A part B wants us to find (rate of change from x = 2 to x = 3)/(rate of change from x = 1 to x = 2) using what we calculated earlier... 100/4 = ?

  43. anonymous
    • one year ago
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    25

  44. Vocaloid
    • one year ago
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    right so section B is 25 times as great (24 times greater) as section A

  45. anonymous
    • one year ago
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    So that's the answer

  46. Vocaloid
    • one year ago
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    as for "explain why" I would say something along the lines of "f(x) is an increasing exponential function

  47. Vocaloid
    • one year ago
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    yeah

  48. anonymous
    • one year ago
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    Thank you

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