/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 Simplify (a) \(\frac{5 z}{z}\), ... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 3

Simplify (a) \(\frac{5 z}{z}\), (b) \(\frac{25 z}{5 z},(c)\) (c) \(\frac{5}{25 z^{2}}\), (d) \(\frac{5 z}{25 z^{2}}\)

Short Answer

Expert verified
(a) 5, (b) 5, (c) \( \frac{1}{5z^2} \), (d) \( \frac{1}{5z} \)

Step by step solution

01

Simplify (a) \( \frac{5z}{z} \)

In the fraction \( \frac{5z}{z} \), both the numerator and denominator have the common factor \( z \). We can cancel the \( z \) in the numerator with the \( z \) in the denominator. This gives us \( 5 \).
02

Simplify (b) \( \frac{25z}{5z} \)

In the fraction \( \frac{25z}{5z} \), we can divide the numerator and the denominator by their common factor, which is \( 5z \). This simplifies to \( \frac{25}{5} = 5 \).
03

Simplify (c) \( \frac{5}{25z^2} \)

For the fraction \( \frac{5}{25z^2} \), the numerator and the denominator can be simplified by their common factor \( 5 \). This results in \( \frac{1}{5z^2} \).
04

Simplify (d) \( \frac{5z}{25z^2} \)

In the fraction \( \frac{5z}{25z^2} \), both numerator and denominator can be simplified by dividing by their common factor \( 5z \). This leaves us with \( \frac{1}{5z} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Common Factors
Finding common factors in algebra is like identifying common ingredients in a recipe.
It simplifies the process and gives you a clearer understanding of the overall problem.
In mathematics, a common factor is a number or variable that divides exactly into two or more numbers.
  • In the expression \( \frac{5z}{z} \), the common factor is \( z \).
  • In \( \frac{25z}{5z} \), the common factor involves \( 5z \).
  • For \( \frac{5}{25z^2} \), the only common factor is \( 5 \).
  • Lastly, in \( \frac{5z}{25z^2} \), the common factor is \( 5z \).
The goal is to make the fraction simpler.
Think of common factors as a tool to reduce complexity.
Being able to spot them quickly is key to tackling algebra problems efficiently.
Fraction Simplification
Fraction simplification makes mathematics cleaner and more readable.
It's like organizing a messy desk; the clutter goes away and everything becomes easy to find.
Simplifying a fraction means finding its simplest form such that the numerator (top number) and the denominator (bottom number) have no other common factors except 1.
  • For \( \frac{5z}{z} \), cancel through common factors to get just \( 5 \).
  • With \( \frac{25z}{5z} \), divide by the common factor \( 5z \) to deduce \( 5 \).
  • For \( \frac{5}{25z^2} \), use the common factor \( 5 \) to arrive at the simplest form \( \frac{1}{5z^2} \).
  • Finally, in \( \frac{5z}{25z^2} \), cancel common factors \( 5z \), leaving \( \frac{1}{5z} \).
Each simplification highlights the essence of the fraction, making it friendlier, like a puzzle piece easily fitting into a larger solution.
Numerator and Denominator Cancellation
Numerator and denominator cancellation is like decluttering your space by removing unnecessary items.
When a factor is present in both the numerator and the denominator, it can be cancelled or reduced.
This principle helps equations appear less cluttered and more direct.
  • In \( \frac{5z}{z} \), the \( z \) cancels out, leaving \( 5 \).
  • For \( \frac{25z}{5z} \), canceling \( 5z \) results in a neat \( 5 \).
  • In \( \frac{5}{25z^2} \), the \( 5 \) in the numerator can be removed to simplify the expression as \( \frac{1}{5z^2} \).
  • In \( \frac{5z}{25z^2} \), both \( 5z \)s simplify directly to \( \frac{1}{5z} \).
This technique is essential for algebra, ensuring expressions are as simplified as possible, much like creating a direct path from one point to another.

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Most popular questions from this chapter

Remove the brackets from the following and simplify the result: (a) \((a+3)(a-5)\) (b) \((2 b+6)(b-7)\) (c) \((3 c+4)(2 c-1)\) (d) \((5 d+8)(-2 d-1)\)

Simplify (a) \(\frac{4 x}{3 x}\), (b (b) \(\frac{15 x}{x^{2}}\), (c) \(\frac{4 s}{s^{3}}\), (d) \(\frac{21 x^{4}}{7 x^{3}}\).

The formula for the volume of a cylinder is \(V=\pi r^{2} h .\) Find \(V\) when \(r=5 \mathrm{~cm}\) and \(h=15 \mathrm{~cm}\)

Evaluate without using a calculator: (a) \(\left(\frac{2}{3}\right)^{2}\) (b) \(\left(\frac{2}{5}\right)^{3}\) (c) \(\left(\frac{1}{2}\right)^{2}\) (d) \(\left(\frac{1}{2}\right)^{3}\) (e) \(0.1^{3}\)

Evaluate, without using a calculator: (a) \(3^{3}\) (b) \(3^{4}\) (c) \(2^{5}\)
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