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2 years ago

6

12) PLS HELP ASAPBRAINLIEST AND 10 + POINTS :)

2 answers:

puteri [66]2 years ago

7 0

Answer:

the first one

Step-by-step explanation:

Scilla [17]2 years ago

3 0

According to the graph the answer is your first choice

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Prove that root 7 is irrational by the method of contradiction

Alchen [17]

Let assume that $\sqrt7$ is a rational number. Therefore it can be expressed as a fraction $\dfrac{a}{b}$ where $a,b\in\mathbb{Z}$ and $\text{gcd}(a,b)=1$ .

$\sqrt7=\dfrac{a}{b}\\\\7=\dfrac{a^2}{b^2}\\\\a^2=7b^2$

This means that $a^2$ is divisible by 7, and therefore also $a$ is divisible by 7.

So, $a=7k$ where $k\in\mathbb{Z}$

$(7k)^2=7b^2\\\\49k^2=7b^2\\\\7k^2=b^2$

Analogically to $a^2=7b^2$ ------- $b^2$ is divisible by 7 and therefore so is $b$ .

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Therefore $\sqrt7$ is an irrational number.

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2 years ago