Answer:
Uninsured motorist bodily injury, or UMBI, pays for medical bills, pain and suffering, lost wages if you can't work after an accident and funeral expenses after a crash with an at-fault driver who doesn't have car insurance. It may also cover you if you're hit as a pedestrian or while riding your bike.
Step-by-step explanation:
The surface area would be 33.6
Lets say W=Wesley, M=Max, and J=Jared.
So your equations are M=2J, 2M=W, and M+J+W=21days(3 weeks)
So substituting M and W into M+J+W=21 gives you 2J+J+2M=21, substitute M again, to get 2J+J+4J=21 so you get 7J=21 so J= 3 days. Using J=3, substitute in to find M, using M=2J, so M=2(3) so M=6 and using that, substitute into 2M=W, so 2(6)=W so W=12. And 12+6+3=21 to check.
Answer:17
Step-by-step explanation:
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
