Answer:
Dom(gof)=Dom(f)={1,3,4}.
(gof)(1)=g{f(1)}=g(2)=3.
(gof)(3)=g{f(3)}=g(5)=1.
(gof)(4)=g{f(4)}=g(1)=3.
∴gof={(1,3),(3,1),(4,3)}.
Answer:
With j = number of jars: 2,578 / 6 = j
Step-by-step explanation:
What is a monomials again?
By elimination:
y = 3x - 1
y = 2x + 2
Subtract the second equation from the first
0 = x - 1
y = 2x + 2
Subtract the first equation from the second
0 = x - 1
y = x + 3
Subtract the first equation from the second again
0 = x - 1
y = 4
Subtract x from both sides of the first equation
- x = - 1
y = 4
Divide the first equation by (-1)
x = 1
y = 4
<h3>
So, the solution is x = 1 and y = 4 {or: (1, 4)}</h3>
<h3>
Answer: 126</h3>
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Work Shown:
Let x and y be the two numbers.
We're given x = 162 and the variable y is unknown.
We're also given LCM = 1134 and HCF = 18
So,
LCM = (x*y)/HCF
1134 = 162*y/18
1134 = (162/18)y
1134 = 9y
9y = 1134
y = 1134/9
y = 126
The other number is 126
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Notice that
showing that 18 is the highest common factor (HCF) of the numbers 162 and 126. This partially confirms the answer.
Now let,
- A = multiples of 162
- B = multiples of 126
So,
- A = 162, 324, 486, 648, 810, 972, 1134, 1296, ...
- B = 126, 252, 378, 504, 630, 756, 882, 1008, 1134, 1260, ...
We see that 1134 is in each list of multiples and the smallest such common item. So the lowest common multiple (LCM) of 162 and 126 is 1134. This helps fully confirm the answer.
