He is now 69 years old because i am smarter than you.
Step-by-step explanation:
So, assuming that x would be the number of miles towed, this word problem gives us two linear equations:
One for Brook's Body Shop, which is y = 3x + 25
And one for Morgan Motors, which is y = 5x
We see that these two lines intersect at the point (12.5, 62.5). Remember that the x-value stands for the number of miles. So, at 12.5 miles, both companies charge the same rate.
This means that Morgan Motors is cheaper for any number of miles less than 12.5. In contrast, Brook's Body Shop is cheaper for any number of miles more than 12.5.
In conclusion, if you had to be towed less than 12.5 miles, you should go to Morgan Motors because they are cheaper for short distances. If you had to be towed more than 12.5 miles, you should go to Brook's Body Shop because they are cheaper for longer distances.
Answer:
-1
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
Answer:
y = 2x - 3
Step-by-step explanation:
