To add these amounts together, we must first find their least common multiple in order to get common denominators (b/c when you add fractions, the denominators must be the same).
We'll start by listing some of their multiples.
To do this, count by whatever the denominator is:
4 1/2 (denominator is 2): 2 4 6 8 10 12 14
2 1/4 (denominator is 4): 4 8 12 16
6 1/3 (denominator is 3): 3 6 9 12 15
Look and see which is the first multiple that all three denominators have. Circle them if it helps you. In this case, it's 12.
So now we have to multiply the denominators by whatever number it takes to reach 12, and multiply by the same number to the numerator:
4 1/2 (times 6 to both top and bottom) =
4 6/12
2 1/4 (times 3) = 2 3/12
6 1/3 (times 4) = 6 4/12
Add all these fractions together, and you get 12 13/12, which is equal to 13 1/12.
Thus, Peter makes a total of 13 1/2 cups.
Hope this made sense! tell me if anything is confusing/incorrect :))
Answer:
The length of the fence needed to surround this garden is 188 meters.
Step-by-step explanation:
Given : A fence is guarding off a vegetable garden in the form of a rectangle. It has one side that is 10 m greater than the other side.
To find : The length of the fence needed to surround this garden if the area of the vegetable garden is 2184 m² ?
Solution :
Let the one side of rectangle be 'x'.
Then the other side is 'x+10'.
The area of the rectangle is 2184 m²,
i.e. 

Solve by middle term split,




Reject negative value,
The side of the rectangle is 42 m.
The other side is 42+10=52 m
The perimeter of the rectangle is 



Therefore, the length of the fence needed to surround this garden is 188 meter.
We will traduce the sentences into equations.
Let x be the number of hours in the first job, and y be the number of hours in the second job.
Then the equations are:

The above system has many solution, we can select for example:
x = 1, y = 16. So we can work one hour at the first job and 16 hour at the second job.
If there is such a scalar function <em>f</em>, then



Integrate both sides of the first equation with respect to <em>x</em> :

Differentiate both sides with respect to <em>y</em> :


Integrate both sides with respect to <em>y</em> :

Plug this into the equation above with <em>f</em> , then differentiate both sides with respect to <em>z</em> :



Integrate both sides with respect to <em>z</em> :

So we end up with
