Question

(i) (3^0+4^-1)×2^2 (ii) (3÷3^10)×3^-5 (iii)2^-3×(-7)-3​

Answer 1

Step-by-step explanation:

(3^0+4^-1)×2^2 (ii) (3÷3^10)×3^-5 (iii)2^-3×(-7)-3

Answer 2

Answer: Let points (−2,3,5),(1,2,3) and (7,0,−1) be denoted by P,Q and R respectively.

Step-by-step explanation: Points P,Q and R are collinear if they lie on a line.

PQ=  

(1+2)  

2

+(2−3)  

2

+(3−5)  

2

 

​

 

=  

(3)  

2

+(−1)  

2

+(−2)  

2

 

​

 

=  

9+1+4

​

 

=  

14

​

 

QR=  

(7−1)  

2

+(0−2)  

2

+(−1−3)  

2

 

​

 

=  

(6)  

2

+(−2)  

2

+(−4)  

2

 

​

 

=  

36+4+16

​

 

=  

56

​

 

=2  

14

​

 

PR=  

(7+2)  

2

+(0−3)  

2

+(−1−5)  

2

 

​

 

=  

(9)  

2

+(−3)  

2

+(−6)  

2

 

​

 

=  

81+9+36

​

 

=  

126

​

 

=3  

14

​

 

Here PQ+QR=  

14

​

+2  

14

​

 

=3  

14

​

 

=PR

⇒PQ+QR=PR

Hence points P(−2,3,5),Q(1,2,3) and R(7,0,−1) are collinear