# Math Help - Radical numbers homework due tomorrow

1. ## Radical numbers homework due tomorrow

(9x)
Would that be 9x

-125c
Would that be -5c

Write ^7x by using rational exponents rather than radical notation.

Use the laws of exponents to simplify. (x^-2/3 y)

Find the domain of the given function. F(x)= 2x-3
The 2x-3 is under the radical sign.

Thanks

2. Originally Posted by rowdy3
(9x)
Would that be 9x
Actually $\pm 9x$.

Originally Posted by rowdy3
-125c
Would that be -5c
Yes.
Originally Posted by rowdy3
Write ^7x by using rational exponents rather than radical notation.
$\sqrt[n]{x} = x^{1/n}$
Originally Posted by rowdy3
Use the laws of exponents to simplify. (x^-2/3 y)
$\left ( x^{-2/3}y \right )^3 = \left ( x^{1/3} \right )^3 y^3$ = ?

Originally Posted by rowdy3
Find the domain of the given function. F(x)= 2x-3
The 2x-3 is under the radical sign.
Can the number under the square root be negative?

-Dan

3. Find the domain of the given function. F(x)= 2x-3
The 2x-3 is under the radical sign.
all real numbers where x is greater than or equal to 3/2

Write ^7x by using rational exponents rather than radical notation.
so would it be x^1/3

(x^-2/3 y)
Would the 3's cancel each other out? Would I get xy^3?

4. Originally Posted by rowdy3
Find the domain of the given function. F(x)= 2x-3
The 2x-3 is under the radical sign.
all real numbers where x is greater than or equal to 3/2
(x^-2/3 y)
Would the 3's cancel each other out? Would I get xy^3?
Yes.

Write ^7x by using rational exponents rather than radical notation.
so would it be x^1/3
Well, if I'm reading this correctly you have
$\sqrt[7]{x^3}$
so what does this become? What does the 7 do? And the 3 isn't a cube root so you don't do a 1/3....

-Dan

5. Originally Posted by topsquark
Actually $\pm 9x$.
Hello,

It depends on where the square is, (9x) or above the square root ?
Actually, whatever... a square number is always positive and unless there is a - sign in front of a square root, it's positive...

Isn't it ? ;s

6. Originally Posted by Moo
Hello,

It depends on where the square is, (9x) or above the square root ?
Actually, whatever... a square number is always positive and unless there is a - sign in front of a square root, it's positive...

Isn't it ? ;s
$\sqrt{9} = 3$ by convention.

$\sqrt{9x^2} = \pm 3x$ because we don't know what x is.

That's the way I learned it, anyway.

But you do have a valid point: $\left ( \sqrt{9x} \right ) ^2 = 9x$, whereas $\sqrt{(9x)^2} = \pm 9x$, so we do need to know where that exponent is in the expression.

-Dan

7. Yep, it has to be the absolute value of x actually, in the two configurations

And this is if we work in $\mathbb{R}$ of course

8. 1. 3 50 - 2 18 + 32

2. 2 48x^5 + 4x 27x

3. 7/5

1. 350 - 218 + 32
= 3(2*25) - 2(2*9) + (2*16)
= 152 -62 + 42
= 132

2.2(48x^5) + 4x(27x)
= 2(3*16x^4*x) + 4x(3*9x)
= 8x(3x) + 12x(3x)
= 20x(3x)
Those boxes are square root signs.

9. 7/5