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babymother [125]
3 years ago
9

(-3 2/3)^2 (thast 3 and 2 thirds btw)

Mathematics
1 answer:
Sophie [7]3 years ago
4 0

Hey there! :)

Answer:

\frac{121}{9}

Step-by-step explanation:

We can solve this by first converting -3 2/3 to an improper fraction:

-3\frac{2}{3}  = \frac{-11}{3}

Square the fraction:

(\frac{-11}{3})^{2} = (\frac{-11}{3}) * (\frac{-11}{3}) = \frac{121}{9}

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7 0
3 years ago
What is the median of 34, 16, 41, 20, 56, 81, 62, 74, 62, 12, 22, 50
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Step-by-step explanation:

7 0
3 years ago
This is from Khan academy I have to attach a PNG if you can help me solve it! Thank you!
Mekhanik [1.2K]

49^2m-m : Not equivalent

7^2m-2m : Not equivalent

7^2m-m : Not equivalent

This is actually a trick question. All of the following are actually false statements. Want to know why? Let me show you.

For exponents, if you are dividing a number to some power (i.e 5^3) by the SAME number to a different power (i.e 5^2), then the expression is 5^3-2 or 5^1 = 5. This is true for any number a such that a^b ÷ a^c = a^b-c.

Since 7 and 49 are not the same number, this rule does not apply and thus cannot be simplified any further.

Let me prove why. 5^3 = 125, and 5^2 = 25, and 125 ÷ 25 = 5. This is also the same as 5^3-2 = 5^1 = 5. We just proved this as so.

But, what about 7 and 49, or 2 different numbers. Well it doesn't apply. 7^3 = 343, and 49^2 = 2401, and 343 ÷ 2401 ≈ 0.14. Thus, this is NOT equal to 7^3-2 which is 7. We just proved that a^y ÷ b^z ≠ a^y-z. Congratulations!

Hope this helped!

7 0
2 years ago
HELP please n thank you
JulijaS [17]

Answer:What's wrong with this Question ?

Step-by-step explanation:

6 0
3 years ago
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