Answer:
The whole number dimension that would allow the student to maximize the volume while keeping the surface area at most 160 square is 6 ft
Step-by-step explanation:
Here we are required find the size of the sides of a dunk tank (cube with open top) such that the surface area is ≤ 160 ft²
For maximum volume, the side length, s of the cube must all be equal ;
Therefore area of one side = s²
Number of sides in a cube with top open = 5 sides
Area of surface = 5 × s² = 180
Therefore s² = 180/5 = 36
s² = 36
s = √36 = 6 ft
Therefore, the whole number dimension that would allow the student to maximize the volume while keeping the surface area at most 160 square = 6 ft.
Step-by-step explanation:
(T-S)(2000), when T(x) and S(x) are functions, is equivalent to T(2000)-S(2000), which is 17-7=10
We know that T(x) is equal to everything else, and everything else is S(x)+G(x), so T(x)-S(x) = G(x), or dollars spent on general science
Answer: 0,13/3
Step-by-step explanation: Hope this helps - Lily ^_^
Answer:
The correct answer is option D, 20ft
Step-by-step explanation:
After drawing the picture below, you can just use pythagorean theorem to solve for the diagonal, or the hypotenuse in this case;
(Length)² + (Width)² = (Diagonal)²
(16)² + (12)² = (Diagonal)²
256 + 144 = (Diagonal)²
400 = (Diagonal)²
= 
<em>20 = Diagonal</em>
<em>Hope this helps!</em>
Alright, so for every addition or subtraction (without the whole thing being in parenthesis and multiplied by something) sign, there is 1 term to the right of it and 1 term to the left of it. However, we have to make sure not to overlap - for example, m is only 1 term - it can't be counted as 2 since it's to the right of a - and to the left of a +. Our terms are therefore 2m²n, 2mn², m, 3n, and 9
