The slope is 3 y-intercept is 7 and the equation would be y=3x+7
Answer:
$98.4.
Step-by-step explanation:
Since the university has decided that for the following year lecture costs will increase by 20%, workshop costs by 10%, and examination costs will stay the same, and that for level 2 the current hourly costs are 10, 24, and 60 respectively To determine how much the level 2 course cost in the following year, the following calculation must be performed:
10 x 1.2 + 24 x 1.1 + 60 = X
12 + 26.4 + 60 = X
98.4 = X
Thus, the level 2 course will cost $ 98.4 next year.
0.43 is greater because you need 7 to get to 0.5 unlike 0.42 where you need 8
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