QUESTION IMAGE
Question
- while solving the equation 3(x - 2) = 6, becca wrote 3x - 6 = 6. which property did she use?
- distributive
- associative
- commutative
- identity
- when solving the equation 4x² - 5 = 25, laura wrote 4x² = 30, as her first step. which property justifies lauras first step?
- distributive property of multiplication over addition
- multiplication property of equality
- commutative property of addition
- addition property of equality
- given the following expressions circle the ones that are irrational.
√2·√32 √4 + π 1/2+√9/4 (2√3)² (2√3)³ 1/3
explain why these expressions are irrational.
Step1: Analyze problem 6
Becca went from $3(x - 2)$ to $3x-6$. The distributive property $a(b - c)=ab - ac$ was used where $a = 3$, $b=x$ and $c = 2$.
Step2: Analyze problem 7
Laura went from $4x^{2}-5 = 25$ to $4x^{2}=30$. She added 5 to both sides of the equation. The addition property of equality states that if $a=b$, then $a + c=b + c$. Here $a = 4x^{2}-5$, $b = 25$ and $c = 5$.
Step3: Analyze problem 8 - Simplify expressions
- For $\sqrt{2}\cdot\sqrt{32}=\sqrt{2\times32}=\sqrt{64}=8$ (rational).
- For $\sqrt{4}+\pi=2 + \pi$. Since $\pi$ is irrational, $2+\pi$ is irrational.
- For $\frac{1}{2}+\sqrt{\frac{9}{4}}=\frac{1}{2}+\frac{3}{2}=2$ (rational).
- For $(2\sqrt{3})^{2}=2^{2}\times(\sqrt{3})^{2}=4\times3 = 12$ (rational).
- For $(2\sqrt{3})^{3}=2^{3}\times(\sqrt{3})^{3}=8\times3\sqrt{3}=24\sqrt{3}$. Since $\sqrt{3}$ is irrational, $24\sqrt{3}$ is irrational.
- For $\frac{1}{3}$ (rational).
$\sqrt{4}+\pi$ is irrational because $\pi$ is irrational and the sum of a rational number ($\sqrt{4}=2$) and an irrational number is irrational. $(2\sqrt{3})^{3}=24\sqrt{3}$ is irrational because $\sqrt{3}$ is irrational and multiplying a non - zero rational number ($24$) by an irrational number gives an irrational number.
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- 1) distributive
- 4) addition property of equality
- Irrational expressions: $\sqrt{4}+\pi$, $(2\sqrt{3})^{3}$