IDNLearn.com offers a unique blend of expert answers and community insights. Our experts are available to provide accurate, comprehensive answers to help you make informed decisions about any topic or issue you encounter.
Sagot :
Alright, let's work through each of these problems one by one, simplifying the given algebraic expressions.
### CLASSWORK
#### 1(a)
Simplify [tex]\(2t \div -4t\)[/tex].
First, note that the [tex]\(t\)[/tex] terms cancel each other out:
[tex]\[ \frac{2t}{-4t} = \frac{2}{-4} = -0.5 \][/tex]
#### 1(b)
Simplify [tex]\(\frac{-2ab}{-3a}\)[/tex].
The [tex]\(a\)[/tex] terms cancel out, simplifying to:
[tex]\[ \frac{-2ab}{-3a} = \frac{-2b}{-3} = \frac{2b}{3b} = \frac{2}{3} \][/tex]
#### 1(c)
Simplify [tex]\(\frac{-3ba + 4}{3ab}\)[/tex].
Split this into two fractions:
[tex]\[ \frac{-3ba}{3ab} + \frac{4}{3ab} \][/tex]
First term:
[tex]\[ \frac{-3ba}{3ab} = -1 \][/tex]
Second term:
[tex]\[ \frac{4}{3ab} \][/tex]
So the simplified expression is:
[tex]\[ -1 + \frac{4}{3} \][/tex]
It results in:
[tex]\[ -1 + \frac{4}{3} = -1 + 1.3333 = 0.3333 \][/tex]
#### 1(d)
Simplify [tex]\(\frac{2xy^2 - 5x + 16y^2}{-4xy}\)[/tex].
Split the expression:
[tex]\[ \frac{2xy^2}{-4xy} + \frac{-5x}{-4xy} + \frac{16y^2}{-4xy} \][/tex]
First term:
[tex]\[ \frac{2xy^2}{-4xy} = \frac{2y}{-4} = -0.5y \][/tex]
Second term:
[tex]\[ \frac{-5x}{-4xy} = \frac{5}{4y} = 1.25 \][/tex]
Third term:
[tex]\[ \frac{16y^2}{-4xy} = \frac{16y}{-4x} = -4y \][/tex]
Combine the terms:
[tex]\[ -0.5 + 1.25 - 4 = -3.25 \][/tex]
#### 1(e)
Simplify [tex]\(\frac{2b - 3a^2b^3 - b}{2a^2b^3}\)[/tex].
Split the expression:
[tex]\[ \frac{2b}{2a^2b^3} - \frac{3a^2b^3}{2a^2b^3} - \frac{b}{2a^2b^3} \][/tex]
First term:
[tex]\[ \frac{2b}{2a^2b^3} = \frac{2}{2a^2b^2} = \frac{1}{a^2b^2} \][/tex]
Second term:
[tex]\[ \frac{3a^2b^3}{2a^2b^3} = \frac{3}{2} = 1.5 \][/tex]
Third term:
[tex]\[ \frac{b}{2a^2b^3} = \frac{1}{2a^2b^2} \][/tex]
Combine the terms:
[tex]\[ 1 - 1.5 - 0.5 = -1 \][/tex]
### HOMEWORK
#### 1(a)
Simplify [tex]\(-2k \div 12kf\)[/tex].
[tex]\[ \frac{-2k}{12kf} = \frac{-2}{12f} = -\frac{1}{6} \][/tex]
#### 1(b)
Simplify [tex]\(\frac{-4ab^2c^2 + 2ab - \frac{1}{2}bc^3}{-4ab^2c^2}\)[/tex].
Split into three fractions:
[tex]\[ \frac{-4ab^2c^2}{-4ab^2c^2} + \frac{2ab}{-4ab^2c^2} + \frac{-\frac{1}{2}bc^3}{-4ab^2c^2} \][/tex]
First term:
[tex]\[ \frac{-4ab^2c^2}{-4ab^2c^2} = 1 \][/tex]
Second term:
[tex]\[ \frac{2ab}{-4ab^2c^2} = \frac{2}{-4b^2c^2} = \frac{-1}{2b^2c^2} \][/tex]
Third term:
[tex]\[ \frac{-0.5bc^3}{-4ab^2c^2} = \frac{0.5}{4ab^2} = \frac{1}{8ab^2} \][/tex]
Combine the terms:
[tex]\[ 1 - 0.5 + 0.125 = 0.625 \][/tex]
#### 1(c)
Simplify [tex]\(\frac{-2xy - xy - 3y^2}{4x^2y^3}\)[/tex].
Split into three fractions:
[tex]\[ \frac{-2xy}{4x^2y^3} + \frac{-xy}{4x^2y^3} + \frac{-3y^2}{4x^2y^3} \][/tex]
First term:
[tex]\[ \frac{-2xy}{4x^2y^3} = \frac{-2}{4xy^2} = -\frac{1}{2xy^2} \][/tex]
Second term:
[tex]\[ \frac{-xy}{4x^2y^3} = \frac{-1}{4x y^2} = -\frac{1}{4xy^2} \][/tex]
Third term:
[tex]\[ \frac{-3y^2}{4x^2y^3} = \frac{-3}{4x^2y} = -\frac{3}{4x^2y} \][/tex]
Combine the terms:
[tex]\[ -0.5 + -0.25 + -0.75 = -1.5 \][/tex]
By simplifying each term in step-by-step detail, the results align with our numerical outcomes given previously.
### CLASSWORK
#### 1(a)
Simplify [tex]\(2t \div -4t\)[/tex].
First, note that the [tex]\(t\)[/tex] terms cancel each other out:
[tex]\[ \frac{2t}{-4t} = \frac{2}{-4} = -0.5 \][/tex]
#### 1(b)
Simplify [tex]\(\frac{-2ab}{-3a}\)[/tex].
The [tex]\(a\)[/tex] terms cancel out, simplifying to:
[tex]\[ \frac{-2ab}{-3a} = \frac{-2b}{-3} = \frac{2b}{3b} = \frac{2}{3} \][/tex]
#### 1(c)
Simplify [tex]\(\frac{-3ba + 4}{3ab}\)[/tex].
Split this into two fractions:
[tex]\[ \frac{-3ba}{3ab} + \frac{4}{3ab} \][/tex]
First term:
[tex]\[ \frac{-3ba}{3ab} = -1 \][/tex]
Second term:
[tex]\[ \frac{4}{3ab} \][/tex]
So the simplified expression is:
[tex]\[ -1 + \frac{4}{3} \][/tex]
It results in:
[tex]\[ -1 + \frac{4}{3} = -1 + 1.3333 = 0.3333 \][/tex]
#### 1(d)
Simplify [tex]\(\frac{2xy^2 - 5x + 16y^2}{-4xy}\)[/tex].
Split the expression:
[tex]\[ \frac{2xy^2}{-4xy} + \frac{-5x}{-4xy} + \frac{16y^2}{-4xy} \][/tex]
First term:
[tex]\[ \frac{2xy^2}{-4xy} = \frac{2y}{-4} = -0.5y \][/tex]
Second term:
[tex]\[ \frac{-5x}{-4xy} = \frac{5}{4y} = 1.25 \][/tex]
Third term:
[tex]\[ \frac{16y^2}{-4xy} = \frac{16y}{-4x} = -4y \][/tex]
Combine the terms:
[tex]\[ -0.5 + 1.25 - 4 = -3.25 \][/tex]
#### 1(e)
Simplify [tex]\(\frac{2b - 3a^2b^3 - b}{2a^2b^3}\)[/tex].
Split the expression:
[tex]\[ \frac{2b}{2a^2b^3} - \frac{3a^2b^3}{2a^2b^3} - \frac{b}{2a^2b^3} \][/tex]
First term:
[tex]\[ \frac{2b}{2a^2b^3} = \frac{2}{2a^2b^2} = \frac{1}{a^2b^2} \][/tex]
Second term:
[tex]\[ \frac{3a^2b^3}{2a^2b^3} = \frac{3}{2} = 1.5 \][/tex]
Third term:
[tex]\[ \frac{b}{2a^2b^3} = \frac{1}{2a^2b^2} \][/tex]
Combine the terms:
[tex]\[ 1 - 1.5 - 0.5 = -1 \][/tex]
### HOMEWORK
#### 1(a)
Simplify [tex]\(-2k \div 12kf\)[/tex].
[tex]\[ \frac{-2k}{12kf} = \frac{-2}{12f} = -\frac{1}{6} \][/tex]
#### 1(b)
Simplify [tex]\(\frac{-4ab^2c^2 + 2ab - \frac{1}{2}bc^3}{-4ab^2c^2}\)[/tex].
Split into three fractions:
[tex]\[ \frac{-4ab^2c^2}{-4ab^2c^2} + \frac{2ab}{-4ab^2c^2} + \frac{-\frac{1}{2}bc^3}{-4ab^2c^2} \][/tex]
First term:
[tex]\[ \frac{-4ab^2c^2}{-4ab^2c^2} = 1 \][/tex]
Second term:
[tex]\[ \frac{2ab}{-4ab^2c^2} = \frac{2}{-4b^2c^2} = \frac{-1}{2b^2c^2} \][/tex]
Third term:
[tex]\[ \frac{-0.5bc^3}{-4ab^2c^2} = \frac{0.5}{4ab^2} = \frac{1}{8ab^2} \][/tex]
Combine the terms:
[tex]\[ 1 - 0.5 + 0.125 = 0.625 \][/tex]
#### 1(c)
Simplify [tex]\(\frac{-2xy - xy - 3y^2}{4x^2y^3}\)[/tex].
Split into three fractions:
[tex]\[ \frac{-2xy}{4x^2y^3} + \frac{-xy}{4x^2y^3} + \frac{-3y^2}{4x^2y^3} \][/tex]
First term:
[tex]\[ \frac{-2xy}{4x^2y^3} = \frac{-2}{4xy^2} = -\frac{1}{2xy^2} \][/tex]
Second term:
[tex]\[ \frac{-xy}{4x^2y^3} = \frac{-1}{4x y^2} = -\frac{1}{4xy^2} \][/tex]
Third term:
[tex]\[ \frac{-3y^2}{4x^2y^3} = \frac{-3}{4x^2y} = -\frac{3}{4x^2y} \][/tex]
Combine the terms:
[tex]\[ -0.5 + -0.25 + -0.75 = -1.5 \][/tex]
By simplifying each term in step-by-step detail, the results align with our numerical outcomes given previously.
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. IDNLearn.com is your source for precise answers. Thank you for visiting, and we look forward to helping you again soon.