True or False

antoniya [11.8K]

Market 1: 3,90/10 = 0.390 $
market 2: 4,44/12= 0,37 $
best price in market 2

4 0

3 years ago

find the answer to start fraction square root of 196 end square root over seven end fraction times square root of 108 end square

mel-nik [20]

1. From your description, I can infer that the multiplication is:
$\frac{ \sqrt{196} }{7} * \sqrt{108}$

The first thing we are going to do is simplify the radicands 196 ans 108 (picture 1):
$196=2^2*7^2$ and $108=2^2*3^3$
Knowing this, we can rewrite our radicals as follows:
$\frac{ \sqrt{196} }{7} * \sqrt{108}= \frac{ \sqrt{2^2*7^2} }{7} * \sqrt{2^2*3^3}$

Remember that $\sqrt[n]{x^n} =x$ ; in other words if the radicand is raised to the same power as the index of the radical, we can take the radicand out. Since 2 and 7 are raised to the power 2 and the index of the radical is also 2 (square root), we can take out 2 and 7:
$\frac{ \sqrt{2^2*7^2} }{7} * \sqrt{2^2*3^3}= \frac{2*7}{7} *2 \sqrt{3^3}$

Look! we have the same numerator and denominator in our fraction, so we can cancel them both:
$\frac{2*7}{7} *2 \sqrt{3^3}=2*2 \sqrt{3^3} =4 \sqrt{3^3}$

Notice that we can write $3^3$ as $3^2*3$ , so we can rewrite our expression one last time:
$4 \sqrt{3^3} =4 \sqrt{3^2*3} =4*3 \sqrt{3} =12 \sqrt{3}$

We can conclude that the correct option is: $12 \sqrt{3}$

2. The <span>product of a nonzero rational number and an irrational number is always an irrational number.

Proof by contradiction:
Lets assume that the product of an irrational number and a rational non-zero number is always rational.
Let </span> $x$ be and irrational number and let $\frac{a}{b}$ and $\frac{c}{d}$ be two rational numbers with $a$ , $b$ , $c$ , and $d$ are non-zero integers.
$x* \frac{a}{b} = \frac{c}{d}$
$x= \frac{c}{d} * \frac{b}{a}$
$x=\frac{cb}{da}$
Since integers are closed under multiplication, $\frac{cb}{da}$ is a rational number. Since $x$ is an irrational number and $x=\frac{cb}{da}$ , we have a logical contradiction, so we can conclude that the product of an irrational number and a rational non-zero number is always an irrational number.