Answer:
x | y
0 | 0
6 | 2
12 | 4
Step-by-step explanation:
Multiply each x value in the table by 1/3 to get 0, 2, and 4 for your y values.
Answer:
30
Step-by-step explanation:
Find the lowest number divisible by both 6, 5, and 2.
To start, let's list off the numbers divisible by six, and see if we can check any of them off for 5. Since both 2 and 6 are even numbers, we know that if a number is divisible by 6, it's divisible by 2.
6 12 18 24 30 36 42 48 54 60
All of these numbers are divisible by two. Let's find the lowest one that is divisible by 5. We know this by either the umber ending in 5 or 0.
30 is the lowest number that is divisible by 2, 5, or 6.
Answer:
x = 48 and y = 104
Step-by-step explanation:
Given equations are:

From equation 1:

Putting the value of y in equation 2

Now we have to put the value of y in one of the equation to find the value of x
Putting y = 104 in the first equation

Hence,
The solution of the system of equations is x = 48 and y = 104
The value of variable which was assumed for number of hats, is the total number of hats.
Given function:

The minimum value of the function can be found by setting the first derivative of the function to zero.


Solving for x:


Substituting the value of x into the original function:

Hence, the minimum value in the given range is (-1, -0.368)
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