8/100, 3/5, 7/10
Explanation:
8/100 = 0.08
3/5 = 0.6
7/10 = 0.7
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
Divide the top and bottom by 3 (since it is the GCF of 3 and 9) to reduce the fraction
3/3=1
9/3=3
answer: 1/3
Cos(225)= -0.71 (2.dp)
tan(-150)= 0.58 (2.dp)
Answer:
C. (1,-1)
Step-by-step explanation:
Lines are intersecting each other at point (1, - 1).
Hence, the system of equations graphed here would be (1, - 1)