do these problems on some clean paper. label each page of your work with your name, your class, the date and…

do these problems on some clean paper. label each page of your work with your name, your class, the date and the book number. also number each problem. keep this written work inside your book, and turn it in with your book when you are finished. please do a neat job. 1. explain why 1.4142135 ≈ √2. 2. copy each expression which stands for a rational number and tell what rational number it equals. √5 √25 √20 √20√5 √5√5 √20/√5 √25/√5 √25/√4 -√5 -√4 √-4 √-1 √0 √1 √10 √100 √1000 √10000 3. simplify each expression. √50 √3 √(1/7) √(5/9) √(5/6) 2√12 3/√2 √6x² √49x √24 + √600

do these problems on some clean paper. label each page of your work with your name, your class, the date and the book number. also number each problem. keep this written work inside your book, and turn it in with your book when you are finished. please do a neat job. 1. explain why 1.4142135 ≈ √2. 2. copy each expression which stands for a rational number and tell what rational number it equals. √5 √25 √20 √20√5 √5√5 √20/√5 √25/√5 √25/√4 -√5 -√4 √-4 √-1 √0 √1 √10 √100 √1000 √10000 3. simplify each expression. √50 √3 √(1/7) √(5/9) √(5/6) 2√12 3/√2 √6x² √49x √24 + √600

Answer

1.

Answer:

The square - root of 2 ($\sqrt{2}$) is an irrational number. When we calculate its decimal approximation, we find that $\sqrt{2}=1.414213562373095\cdots$. Rounding this value to seven decimal places gives 1.4142135, which is why $1.4142135\approx\sqrt{2}$.

Explanation:

Step1: Recall the definition of $\sqrt{2}$

$\sqrt{2}$ is the positive number that, when multiplied by itself, gives 2.

Step2: Calculate the decimal expansion

Using a calculator or long - division method for square roots, we find $\sqrt{2}=1.414213562373095\cdots$.

Step3: Round the decimal

Rounding to seven decimal places, we get 1.4142135.

2.

Answer:

  • $\sqrt{5}$ is irrational.
  • $\sqrt{25}=5$ (since $5\times5 = 25$).
  • $\sqrt{20}=2\sqrt{5}$ is irrational.
  • $\sqrt{20}\sqrt{5}=\sqrt{100}=10$ (using $\sqrt{a}\sqrt{b}=\sqrt{ab}$).
  • $\sqrt{5}\sqrt{5}=\sqrt{25}=5$.
  • $\frac{\sqrt{20}}{\sqrt{5}}=\sqrt{\frac{20}{5}}=\sqrt{4}=2$.
  • $\frac{\sqrt{25}}{\sqrt{5}}=\sqrt{\frac{25}{5}}=\sqrt{5}$ is irrational.
  • $\frac{\sqrt{25}}{\sqrt{4}}=\frac{5}{2}=2.5$.
  • $-\sqrt{5}$ is irrational.
  • $-\sqrt{4}=- 2$.
  • $\sqrt{-4}$ and $\sqrt{-1}$ are non - real (complex numbers).
  • $\sqrt{0}=0$.
  • $\sqrt{1}=1$.
  • $\sqrt{10}$ is irrational.
  • $\sqrt{100}=10$.
  • $\sqrt{1000}=10\sqrt{10}$ is irrational.
  • $\sqrt{10000}=100$.

Explanation:

Step1: Recall the definition of square root

If $x^{2}=a$, then $\sqrt{a}=x$ (for $a\geq0$ and $x\geq0$).

Step2: Use square - root properties

For $\sqrt{a}\sqrt{b}=\sqrt{ab}$ and $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}(a\geq0,b > 0)$.

Step3: Identify rational and irrational numbers

Rational numbers can be written as $\frac{p}{q}$ where $p,q\in\mathbb{Z}$ and $q\neq0$. Irrational numbers cannot be written in this form.

3.

Answer:

  • $\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}$.
  • $\sqrt{3}$ is already in simplest form.
  • $\sqrt{\frac{1}{7}}=\frac{\sqrt{1}}{\sqrt{7}}=\frac{1}{\sqrt{7}}=\frac{\sqrt{7}}{7}$.
  • $\sqrt{\frac{5}{9}}=\frac{\sqrt{5}}{\sqrt{9}}=\frac{\sqrt{5}}{3}$.
  • $\sqrt{\frac{5}{6}}=\frac{\sqrt{5}}{\sqrt{6}}=\frac{\sqrt{5}\times\sqrt{6}}{6}=\frac{\sqrt{30}}{6}$.
  • $2\sqrt{12}=2\sqrt{4\times3}=4\sqrt{3}$.
  • $\frac{3}{\sqrt{2}}=\frac{3\sqrt{2}}{2}$.
  • $\sqrt{6x^{2}}=\vert x\vert\sqrt{6}$.
  • $\sqrt{49x}=7\sqrt{x}(x\geq0)$.
  • $\sqrt{24}+\sqrt{600}=\sqrt{4\times6}+\sqrt{100\times6}=2\sqrt{6}+10\sqrt{6}=12\sqrt{6}$.

Explanation:

Step1: Factor the numbers inside the square root

For example, $50 = 25\times2$, $12 = 4\times3$, $24=4\times6$, $600 = 100\times6$.

Step2: Use the property $\sqrt{ab}=\sqrt{a}\sqrt{b}(a\geq0,b\geq0)$

$\sqrt{25\times2}=\sqrt{25}\sqrt{2}=5\sqrt{2}$.

Step3: Rationalize the denominator

For $\frac{3}{\sqrt{2}}$, multiply numerator and denominator by $\sqrt{2}$ to get $\frac{3\sqrt{2}}{2}$.

Step4: Simplify expressions with variables

For $\sqrt{6x^{2}}$, since $\sqrt{x^{2}}=\vert x\vert$, we have $\sqrt{6x^{2}}=\vert x\vert\sqrt{6}$.