Question

find minus 1 by 6 x 4 by 7 plus one by two - 3 by 7 x 1 by 6​

Answer 1

Step-by-step explanation:

(i) To verify : A×(B∩C)=(A×B)∩(A×C)

We have B∩C={1,2,3,4}∩{5,6}=ϕ

∴ L.H.S = A×(B∩C)=A×ϕ=ϕ

A×B={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}

A×C={(1,5),(1,6),(2,5),(2,6)}

∴R.H.S.=(A×B)∩(A×C)=ϕ

∴L.H.S=R.H.S

Hence A×(B∩C)=(A×B)∩(A×C)

(ii) To verify: A×C is a subset of B×D

A×C={(1,5),(1,6),(2,5),(2,6)}

B×D={(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7),

(3,8),(4,5),(4,6),(4,7),(4,8)}

We can observe that all the elements of set A×C are the elements of set B×D

Therefore A×C is a subset of B×D

Answer 2

Step-by-step explanation:

Given: −3(x−7)≥15−7x>

3

x+1

−3x+21≥15−7x>

3

x+1

⇒−3x+21≥15−7x and 155−7x>

3

x+1

⇒4x≥−6 and 45−21x>x+1

⇒x≥−

2

3

and 44>22x

⇒x≥

2

−3

and 2>x

∴ Solution set ={x:x∈R,

2

−3

≤x<2}