The given equations are linear functions and represent straight lines.
The first equation F(x) = 2x , represent a straight line passing through the origin.
Now, we plot the rest equation using the transformation of the graph.
- If we add a constant 'a' in the function then it gets shifted upward by 'a' units.
- If we subtract a constant 'a' in the function then it gets shifted downward by 'a' units.
Using these properties, the second graph gets shifted upward by 6 units. Hence, it passes through (0,6)
Third graph will shift downward by 3 units. Hence, it passes through the point (0,-3)
Fourth graph will shift downward by 5 units. Hence, it passes through the point (0,-5)
The graph of these functions are shown in the same xy-plane in the attached file.
Answer:
The dimensions that minimize the surface are:
Wide: 1.65 yd
Long: 3.30 yd
Height: 2.20 yd
Step-by-step explanation:
We have a rectangular base, that its twice as long as it is wide.
It must hold 12 yd^3 of debris.
We have to minimize the surface area, subjet to the restriction of volume (12 yd^3).
The surface is equal to:
The volume restriction is:
If we replace h in the surface equation, we have:
To optimize, we derive and equal to zero:
![dS/dw=36(-1)w^{-2} + 8w=0\\\\36w^{-2}=8w\\\\w^3=36/8=4.5\\\\w=\sqrt[3]{4.5} =1.65](https://tex.z-dn.net/?f=dS%2Fdw%3D36%28-1%29w%5E%7B-2%7D%20%2B%208w%3D0%5C%5C%5C%5C36w%5E%7B-2%7D%3D8w%5C%5C%5C%5Cw%5E3%3D36%2F8%3D4.5%5C%5C%5C%5Cw%3D%5Csqrt%5B3%5D%7B4.5%7D%20%3D1.65)
Then, the height h is:
The dimensions that minimize the surface are:
Wide: 1.65 yd
Long: 3.30 yd
Height: 2.20 yd
Answer:
This is a postulate which states that through any two points, there is exactly one line.
Step-by-step explanation:
A postulate is a statement that is assumed true without proof.
Answer:
The distance between the two playing pieces is of 6.4 units.
Step-by-step explanation:
Suppose we have two points:
The distance between these points is:
Distance between:
A(-2,3) and B(3,-1)
The distance between the two playing pieces is of 6.4 units.
