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Papessa [141]
3 years ago
9

Rewrite the expression with rational exponents as a radical expression by extending the properties of integer exponents.

Mathematics
2 answers:
VashaNatasha [74]3 years ago
4 0

Answer:

<u>3^1/3</u>

3^1/6  

I believe this would be correct hope this helps.

-Que

Nikitich [7]3 years ago
4 0

Answer:

\frac{3^{\frac{1}{3} } }{3^{\frac{1}{6} }}

Step-by-step explanation:

Three to the one third power all over three to the one sixth power.

One third is 1/3 and one sixth is 1/6. All over means divided by. The expression before the words "all over" are numerator and the expression after "all over" are denominator.

This becomes:;

\frac{3^{\frac{1}{3} } }{3^{\frac{1}{6} }}

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Answer:

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2 years ago
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Which property was used to simplify the expression?<br><br> (y^5)^2 = y^10
gayaneshka [121]
<span>(y^5)^2 = y^10

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3 0
3 years ago
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Work out the surface area of this cylinder.<br> Radius= 12 cm<br> Height= 35 cm
Ad libitum [116K]
3543.72 cm^2.

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7 0
3 years ago
2x+3y=15 x+y=6 find values for x and y
MatroZZZ [7]

Answer:

  (x, y) = (3, 3)

Step-by-step explanation:

Subtract double the second equation from the first.

  (2x +3y) -2(x +y) = (15) -2(6)

  y = 3 . . . . . . . . simplify

Use the second equation to find x.

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The values of x and y are both 3.

8 0
2 years ago
True or false: sec^-1(0.5) is undefined. explain your answer
ICE Princess25 [194]

Here we get the statement:

"sec^-1(0.5) is undefined"

And we want to see if this is true or false, so let's use the properties that we know to find that this is false.

Remember that the sec function is defined as:

sec(x) = \frac{1}{cos(x)}

Then we will have:

sec^{-1}(x) = (\frac{1}{cos(x)})^{-1} =  cos(x)

Then is really trivial to see that:

sec^{-1}(0.5) =  cos(0.5) = 0.878

Then we can conclude that the function is not undefined at 0.5, so the statement is false.

Below you can see the graph of the given function, and you will see that it is never undefined.

If you want to learn more, you can read:

brainly.com/question/16453813

6 0
3 years ago
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