Answer:
About 90 seats per section.
hope this helps!
Answer:
Step-by-step explanation:
-5x - 8y = -3
5x + 9y = 3
y = 0
5x + 0 = 3
5x = 3
x = 3/5
(3/5, 0)
Answer:
The perimeter is 38.25 units
Step-by-step explanation:
The perimeter of the tra-pezoid is the distance around it.
The length of the bases can be found using the absolute value method.
|RS|=|20-14|=6 units.
|TQ|=|22-8|=14
Recall the distance formula;
![d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
We use the distance to find the non parallel sides.
units.
and
units.
The perimeter of the tra-pezoid is
=14+10+6+8.25=38.25
Answer
78.75
Step-by-step explanation:
75.00 × 0.05
= 78.75
Answer:
10m x 15m
Step-by-step explanation:
You are given some information.
1. The area of the garden: A₁ = 150m²
2. The area of the path: A₂ = 186m²
3. The width of the path: 3m
If the garden has width w and length l, the area of the garden is:
(1) A₁ = l * w
The area of the path is given by:
(2) A₂ = 3l + 3l + 3w + 3w + 4*3*3 = 6l + 6w + 36
Multiplying (2) with l gives:
(3) A₂l = 6l² + 6lw + 36l
Replacing l*w in (3) with A₁ from (1):
(4) A₂l = 6l² + 6A₁ + 36l
Combining:
(5) 6l² + (36 - A₂)l +6A₁ = 0
Simplifying:
(6) l² - 25l + 150 = 0
This equation can be factored:
(7) (l - 10)*(l - 15) = 0
Solving for l we get 2 solutions:
l₁ = 10, l₂ = 15
Using (1) to find w:
w₁ = 15, w₂ = 10
The two solutions are equivalent. The garden has dimensions 10m and 15m.