Question

Question 1: If A = {1,2,3} and B = {2,4,6}
Find A U B
Find A ∩ B

Answer 1

A ∪ B means put all the elements in both A and B in one set without repeating any elements.

A ∩ B means put the items that appear in both sets in a single set without repeating any elements.

For example:

If

A = {9, 8, 7}
B = {8, 6, 4}

Then
A ∪ B = {4, 6, 7, 8, 9}
A ∩ B = {8}

Answer 2

15 - 3(4 - 2x) - 7(x - 8) = 15 - 12 + 6x - 7x + 56
= 56 + 15 - 12 + 6x - 7x
= 56 + 3 + (6 - 7)x
= 59 - x

Answer 3

Remember, when simplifying algebraic expressions, you'll most likely have to use distributive property first. Afterwards, remember to put like terms next to each other before combing. That means put integers next to each other and put algebraic terms of the same variable and degree next to each other before combining them.

Answer 4

Thank you so much....i have one last one it would be great if you can show me how to do it as well

Simplify the expression: (Answer should not have any negative exponents)

(3x^2)^4y^2x^-2/x^-3y^5

Answer 5

I assume it is this

\[\frac{(3x^2)^4y^2x^{-2}}{x^{-3}y^5}\]

Answer 6

Is that what you are given?

Answer 7

Well after applying a rule such as

\[a^{-b} = \frac{1}{a^b}\]

You can re-write it as:
\[\frac{(3x^2)^4y^2x^{3}}{x^{2}y^5}\]

Answer 8

From there, you can apply the rule

\[(ab)^c = a^cb^c\]

And get

\[\frac{81x^8y^2x^{3}}{x^{2}y^5}\]

Answer 9

Afterwards, you can apply rules of combining the products of exponents:

\(x^8x^3 = x^{8 + 3} = x^{11}\) to get:

\[\frac{81x^{11}y^2}{x^{2}y^5}\]

Answer 10

Lastly you can apply the rule for simplifying exponents of the same variable that exists in both the numerator and denominator of fractions:

If you have \(\dfrac{x^b}{x^a}\) and \(b >a\) then:

\(\dfrac{x^b}{x^a} = x^{b-a}\)

Which, in this case means \(\dfrac{81x^{11}y^2}{x^{2}y^5}\) becomes

\(\dfrac{81x^{11-2}}{y^{5-2}} = \dfrac{81x^{9}}{y^{3}} \)