(i) (3^0+4^-1)×2^2 (ii) (3÷3^10)×3^-5 (iii)2^-3×(-7)-3
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Answered by
1
Step-by-step explanation:
(3^0+4^-1)×2^2 (ii) (3÷3^10)×3^-5 (iii)2^-3×(-7)-3
Answered by
0
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
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