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3 years ago

How to solve n-5*2/3=3/4

1 answer:

5 0

$n-5\cdot\dfrac{2}{3}=\dfrac{3}{4}\\\\n-\dfrac{10}{3}=\dfrac{3}{4}\ \ \ \ |\cdot12\\\\12n-12\cdot\dfrac{10}{3}=12\cdot\dfrac{3}{4}\\\\12n-4\cdot10=3\cdot3\\\\12n-40=9\ \ \ \ |+40\\\\12n=49\ \ \ \ \ |:12\\\\n=\dfrac{49}{12}$

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Prove that root 7 is irrational by the method of contradiction

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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 $k\in\mathbb{Z}$

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

But if both numbers $a$ and $b$ are divisible by 7, then $\text{gcd}(a,b)=7$ which contradicts our earlier assumption that $\text{gcd}(a,b)=1$ .

Therefore $\sqrt7$ is an irrational number.

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It’s c.

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2 years ago

What is 10+√50<br> -----------<br> 5

zubka84 [21]

Answer:

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Step-by-step explanation:

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