Answer:
![12x^{2} (x^{3}-5)^{3} (x^{4}+3)^{5} +20x^{3} (x^{3}-5)^{4} (x^{4}+3)^{4}](https://tex.z-dn.net/?f=12x%5E%7B2%7D%20%28x%5E%7B3%7D-5%29%5E%7B3%7D%20%28x%5E%7B4%7D%2B3%29%5E%7B5%7D%20%2B20x%5E%7B3%7D%20%28x%5E%7B3%7D-5%29%5E%7B4%7D%20%28x%5E%7B4%7D%2B3%29%5E%7B4%7D)
Step-by-step explanation:
Firstly let's find the dimension of this large rectangle:(given)
Area of Rectangle = 660 x 66 =43,560 ft²
And we know that 1 acre = 43,560 ft², then each rectangle has an area of 1 acre & the 20 acres will correspond to 20 x 43560 = 871,200 ft²
We know that the 20 acres form a rectangle. We need to know what is their disposition:
1) We would like to know the layout of the rectangles since we have 4 possibilities FOR THE LAYOUTS
Note that W=66 & L=666 = 43,956 ft²/ unit )
lay out shape could be either:(in ft)
1 W by 20 L (Final shape Linear 66 x 13320 = 879,120) or
2 W by 10 L (Final shape Stacked 132 x 6660 = 879,120) or
4 W by 5 L (Final shape Stacked 264 x 3330 = 879,120) or
2) We would like to know the number of participants so that to allocate equal space as well as the pedestrian lane, if possible, if not we will calculated the reserved space allocated for pedestrian/visitors)
3) Depending on the shape given we will calculate the visitor space & we will deduct it from the total space to distribute the remaining among the exhibitors.
4) (SUGGESTION) Assuming it's linear, we will reserve
20ft x 13320 ft = = 266,400 ft² and the remaining 612,720 ft² for exhibitors
5) Depending on the kind of the exhibition, we will divide the 612,720 ft² accordingly
6) How can we select the space allocated for each exhibitor:
the 617,720 ft² could be written as a product of prime factors:
612720 = 2⁴ x 3² x 5 x 23 x 37
If you chose each space will be185 ft² , then we can accommodate up to 3,312 exhibitors.
Obviously you can choose any multiple of the prime factors to specify the area allocated & to calculate the number of exhibitors accordingly
Answer:
View the graph
Step-by-step explanation:
Given a utility function U(q) = -q²-4q+2 in this case, depending on demand (q), this is function because it is a relationship or correspondence between two magnitudes, so that each value of the first corresponds to a single value of the second (or none); for example to q=-2 U(q) = 6, this means that for a demand (-2) the utility will be maximum (6)
, and to -∞<q<(-4-√24)/2 ∧ (-4+√24)/2<q<∞ the utility will be negative; (-4-√24)/2<q< (-4+√24)/2 the utility will be positive and q=(-4-√24)/2 ∧ q= (-4+√24)/2 there will be no utility
Answer: 4.5 miles
Explanation:
When you draw the situation you find two triangles.
1) Triangle to the east of the helicopter
a) elevation angle from the high school to the helicopter = depression angle from the helicopter to the high school = 20°
b) hypotensue = distance between the high school and the helicopter
c) opposite-leg to angle 20° = heigth of the helicopter
d) adyacent leg to the angle 20° = horizontal distance between the high school and the helicopter = x
2) triangle to the west of the helicopter
a) elevation angle from elementary school to the helicopter = depression angle from helicopter to the elementary school = 62°
b) distance between the helicopter and the elementary school = hypotenuse
c) opposite-leg to angle 62° = height of the helicopter
d) adyacent-leg to angle 62° = horizontal distance between the elementary school and the helicopter = 5 - x
3) tangent ratios
a) triangle with the helicpoter and the high school
tan 20° = Height / x ⇒ height = x tan 20°
b) triangle with the helicopter and the elementary school
tan 62° = Height / (5 - x) ⇒ height = (5 - x) tan 62°
c) equal the height from both triangles:
x tan 20° = (5 - x) tan 62°
x tan 20° = 5 tan 62° - x tan 62°
x tan 20° + x tan 62° = 5 tan 62°
x (tan 20° + tan 62°) = 5 tan 62°
⇒ x = 5 tant 62° / ( tan 20° + tan 62°)
⇒ x = 4,19 miles
=> height = x tan 20° = 4,19 tan 20° = 1,525 miles
4) Calculate the hypotenuse of this triangle:
hipotenuese ² = x² + height ² = (4.19)² + (1.525)² = 19.88 miles²
hipotenuse = 4.46 miles
Rounded to the nearest tenth = 4.5 miles
That is the distance between the helicopter and the high school.