Question

plz help me to and these questions
wit steps

Answer 1

1. √20 * √5=√20*5=√100=10
3.13^1/5÷13^1/3=13^1/5-1/3=13^-2/15
5.1/2
6.d
13.b

Answer 2

1.   $\sqrt{20}* \sqrt{5}=\sqrt{20*5}=\sqrt{100}=10$

2. $(125)^{-\frac{1}{3}}=\frac{1}{(125)^\frac{1}{3}}=\frac{1}{(5^3)^\frac{1}{3}}=\frac{1}{5^{3*1/3}}=\frac{1}{5^1}$

3.$\frac{13^{\frac{1}{5}}}{13^{\frac{1}{3}}}=(13)^{\frac{1}{5}-\frac{1}{3}}=(13)^{-\frac{2}{15}}=\frac{1}{(13)^{\frac{2}{15}}}$

4. 
$\sqrt[3]{\frac{54}{250}}= \sqrt[3]{\frac{2*3*3*3}{5*5*5*2}}= \sqrt[3]{\frac{3^3}{5^3}}=\frac{3}{5}$

5.
square root 2 is a constant and is not a variable in algebra like x, y, z...
so √2  =  √2 x⁰
  So it is a polynomial of degree 0.
   for example,    a x + b y is a polynomial of degree 1.

6.  square root 2  is a non terminating noon recurring decimal number.
   it is an irrational number.

7. 
$\sqrt{(3^{-2})}=(3^{-2})^{\frac{1}{2}}=3^{-1}=1/3$

8.
$(-2 - \sqrt{3}(-2+\sqrt3)=(-2)^2-(\sqrt3)^2=4-3=1.$
we used the law:  (a - b)(a + b) = a^2 - b^2

9.
rational numbers between 2/3 and 5/3 are: 
     2/3 =  4/6  and 5/ 3  = 10/6
   so in between 4/6 and  10/6    we have  5/6 and 7/6.

10.
  =  5 + √8 + 3 - √2 - √2 + 6
  =  14 + √(4*2)  - 2 √2 
  = 14 + 2√2 - 2√2
  = 14
       positive and rational

11.
5/7 = 0.7142..        5/7 square =   around  0.49...
7/9  = 0.7777...     its square =   between 0.50  to 0.64
   These are much less than 6.  so (b) not answer.
0.75, 0.7512 are  rational numbers although they are  in the range.
 hence  (c) is the option,  it lies in the given range.

12. 
   6 + √27 - 3 - √3 + 1 - 2√3
  = 4 +  3 √3  - √3 - 2 √3 
   = 4
 rational and positive

13.
1/5 and 4/5 . 
  so the denominator is 5...  then the numerators are from 1 to 4.  With the same denominator, the numerators in between 1 and 4  will result in rational numbers required.
   hence, 2/5 an d  3/5.