Answer:
yes
Step-by-step explanation:
because both of them are horizontal straight lines
I think I would first convert it to hours to solve it and then back to minutes at the end. Super strange it’s been asked for in minutes.
So to convert 20 minutes into hours, you need to multiply by 3.
This means you need to do the same to the other side, cause if they’ve travelled 2.5 miles in 20 minutes, they’d travel 3 times that in an hour.
So this is 7.5m/ph
But then they want it in minutes, so I suppose 7.5/60?
So 0.125m/pm?
i)On z, define a∗b=a−b
here aϵz
+
and bϵz
+
i.e.,a and b are positive integers
Let a=2,b=5⇒2∗5=2−5=−3
But −3 is not a positive integer
i.e., −3∈
/
z
+
hence,∗ is not a binary operation.
ii)On Q,define a∗b=ab−1
Check commutative
∗ is commutative if,a∗b=b∗a
a∗b=ab+1;a∗b=ab+1=ab+1
Since a∗b=b∗aforalla,bϵQ
∗ is commutative.
Check associative
∗ is associative if (a∗b)∗c=a∗(b∗c)
(a∗b)∗c=(ab+1)∗c=(ab+1)c+1=abc+c+1
a∗(b∗c)=a∗(bc+1)=a(bc+1)+1=abc+a+1
Since (a∗b)∗c
=a∗(b∗c)
∗ is not an associative binary operation.
iii)On Q,define a∗b=
2
ab
Check commutative
∗ is commutative is a∗b=b∗a
a∗b=
2
ab
b∗a=
2
ba
=
2
ab
a∗b=b∗a∀a,bϵQ
∗ is commutativve.
Check associative
∗ is associative if (a∗b)∗c=a∗(b∗c)
(a∗b)∗c=
2
(
2
ab
)∗c
=
4
abc
(a∗b)∗c=a∗(b∗c)=
2
a×
2
bc
=
4
abc
Since (a∗b)∗c=a∗(b∗c)∀a,b,cϵQ
∗ is an associative binary operation.
iv)On z
+
, define if a∗b=b∗a
a∗b=2
ab
b∗a=2
ba
=2
ab
Since a∗b=b∗a∀a,b,cϵz
+
∗ is commutative.
Check associative.
∗ is associative if $$
(a∗b)∗c=a∗(b∗c)
(a∗b)∗c=(2
ab
)
∗
c=2
2
ab
c
a∗(b∗c)=a∗(2
ab
)=2
a2
bc
Since (a∗b)∗c
=a∗(b∗c)
∗ is not an associative binary operation.
v)On z
+
define a∗b=a
b
a∗b=a
b
,b∗a=b
a
⇒a∗b
=b∗a
∗ is not commutative.
Check associative
∗ is associative if $$
(a∗b)∗c=a∗(b∗c)
(a∗b)∗c=(a
b
)
∗
c=(a
b
)
c
a∗(b∗c)=a∗(2
bc
)=2
a2
bc
eg:−Leta=2,b=3 and c=4
(a∗b)
∗
c=(2∗3)
∗
4=(2
3
)
∗
4=8∗4=8
4
a∗(b∗c)=2
∗
(3∗4)=2
∗
(3
4
)=2∗81=2
81
Since (a∗b)∗c
=a∗(b∗c)
∗ is not an associative binary operation.
vi)On R−{−1}, define a∗b=
b+1
a
Check commutative
∗ is commutative if a∗b=b∗a
a∗b=
b+1
a
b∗a=
a+1
b
Since a∗b
=b∗a
∗ is not commutatie.
Check associative
∗ is associative if (a∗b)∗c=a∗(b∗c)
(a∗b)∗c=(
b+1
a
)
∗
c=
c
b
a
+1
=
c(b+1)
a
a∗(b∗c)=a∗(
c+1
b
)=
c+1
b
a
=
b
a(c+1)
Since (a∗b)∗c
=a∗(b∗c)
∗ is not a associative binary operation
A. terminating
b. repeating
c. repeating
d. terminating
e. repeating
i think
Answer:
The answer is 100ft2 I think
