Question

Halp!

Answer 1

Part A: Use the properties of exponents to explain why 8 raised to the power of 1 over 3 is called the cube root of 8. (5 points)

Part B: The length of a rectangle is 3 units and its width is square root of 3 unit. Is the area of the rectangle rational or irrational? Justify your answer. (5 points)

Answer 2

A
\[X^{1/y}=\sqrt[y]{X}\]
so \[8^{1/3}=\sqrt[3]{8}=2\]
B
\[\sqrt{3}\] is irrational so the area is irrational

Answer 3

for part A i need to explain not show

Answer 4

For part A:
Let x = \(8^{1/3}\). Then,
\(x^{3} = (8^{1/3})^{3} = 8^{3/3} = 8^1 = 8\)
\(x \times x \times x = 8 \) which implies x is the cube root of 8.

Answer 5

A. For this part, we use the property: (x^a)^b = x^(a*b)


Let x = 8^(1/3)

Now raise both sides to the power of 3.

x^3 = (8^(1/3))^3
x^3 = 8^(1/3 * 3)
x^3 = 8^1
x^3 = 8

Since x^3 (x multiplied 3 times by itself) = 8, then x must be cube root of 8
But recall that originally we had x = 8^(1/3)
So 8^(1/3) is the cube root of 8

B.

Area is irrational: 3 * √3 = 3√3 ----> irrational

Answer 6

Can someone maybe help me with my problems? Thanks.

Answer 7

http://openstudy.com/study#/updates/53ae0c29e4b02c45b0156b95

Answer 8

ummmmm ok?

Answer 9

Thank you! :)

Answer 10

\(\Huge\color{cyan}{✯ℋᵅᵖᵖᵞ ℬⁱʳᵗᑋᵈᵃᵞ✯}\)