First it would be helpful to remove the radicals found in the numerator and denominator. We can do this by the property $\sqrt[n]{x^m} = x^{\frac{m}{n}}$ .

This means that our fraction now becomes:

$\dfrac{x^{\frac{3}{4}}}{x^{\frac{2}{3}}}$

Now, we can subtract the exponents of $x$ because of the fact that the numerator and denominator both have a base of $x$ :

$x^{\frac{3}{4} - \frac{2}{3}}$

$x^{\frac{1}{12}}$

Our answer is $\boxed{x^{\frac{1}{12}}}$ .

5 0

3 years ago

Super-yummy soup, inc. wants to paint their entire soup can red, including the top and bottom. if the diameter of the lid is 5 c

Leokris [45]

The total surface area of the soup can would be equivalent to the sum of the surface area of the 2 covers plus the surface area of the side.

Surface area of the 2 covers = 2 π r^2

Surface area of the 2 covers = 2 π (2.5 cm)^2

Surface area of the 2 covers = 39.27 cm^2

Surface area of the side = 2 π r h

Surface area of the side = 2 π (2.5 cm) (9 cm)

Surface area of the side = 141.37 cm^2

Total surface area = 39.27 cm^2 + 141.37 cm^2

<span>Total surface area = 180.64 cm^2</span>

8 0

3 years ago

A rectangular patio has area of 180 square feet. The width of the patio is 3 feet less than the length. Find the length and widt

kondaur [170]

My calculations got to a decimal so it’s about.

6 0

3 years ago

Read 2 more answers

How do I figure out this problem 3x – 11 < 4 or 4x + 9 ≥ 1

serious [3.7K]

You need to find x for both equations first

3x - 11 < 4 4x + 9 ≥ 1
+ 11. +11. -9. -9

3x < 15 4x ≥ -8
/3. /3 /4. /4

X < 5 X ≥ -2

Then you plug in x in the equations
3(5) - 11 < 4. 4(-2) + 9 ≥ 1
15 - 11 < 4 -8 + 9 ≥ 1
4 < 4 1. ≥ 1

False True

8 0

3 years ago

Any two conditions of an object to float on liquid ​

Any two conditions of an object to float on liquid