Name: [tex]$\qquad$[/tex]
Date: [tex]$\qquad$[/tex] Period: [tex]$\qquad$[/tex]

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Simplify: [tex]$\frac{45k^2 - 90k}{54k}$[/tex]

A. [tex]$\frac{7k - 5}{6k}$[/tex]
B. [tex]$\frac{5(k - 2)}{6}$[/tex]
C. [tex]$\frac{6k}{7k - 5}$[/tex]
D. [tex]$\frac{6}{5(k - 2)}$[/tex]



Answer :

To solve the given expression [tex]\(\frac{45k^2 - 90k}{54k}\)[/tex], we need to simplify it step-by-step. Let's go through the process:

1. Factor the numerator:
[tex]\[ 45k^2 - 90k \][/tex]
You can factor out the greatest common factor (GCF) from the terms in the numerator. The GCF of [tex]\(45k^2\)[/tex] and [tex]\(90k\)[/tex] is [tex]\(45k\)[/tex]. So we factor out [tex]\(45k\)[/tex]:
[tex]\[ 45k(k - 2) \][/tex]

2. Rewrite the expression with the factored numerator:
[tex]\[ \frac{45k(k - 2)}{54k} \][/tex]

3. Simplify the fraction:
First, look for common factors in the numerator and the denominator. We see that both the numerator [tex]\(45k(k - 2)\)[/tex] and the denominator [tex]\(54k\)[/tex] have a common factor of [tex]\(9k\)[/tex].

Factor out [tex]\(9k\)[/tex] from both the numerator and the denominator:
[tex]\[ \frac{45k(k - 2)}{54k} = \frac{(5k)(9)(k - 2)}{6(9)k} \][/tex]

4. Simplify further:
[tex]\[ \frac{(5k)(9)(k - 2)}{6(9)k} = \frac{5(k - 2)}{6} \][/tex]
Here, the [tex]\(9\)[/tex] and [tex]\(k\)[/tex] in both the numerator and denominator cancel each other out.

5. Final expression:
[tex]\[ \frac{5(k - 2)}{6} \][/tex]

Thus, the simplified form of the given expression [tex]\(\frac{45k^2 - 90k}{54k}\)[/tex] is [tex]\(\frac{5(k - 2)}{6}\)[/tex].

Therefore, the correct answer is:
[tex]\[ \boxed{B) \frac{5(k-2)}{6}} \][/tex]