Answer:
43.30 km^2
Step-by-step explanation:
Since the shape is a regular polygon, the sides of the triangle are equal (10km) now find the area on the triangle using herons formula
Working:
Side is 10
Perimeter= 30km
Semiperimeter (s) = 15km
Formula= √s(s-a)(s-b)(s-c)
= √15(15-10)(15-10)(15-10)
= √15*5*5*5
=5√5*15
=5√75
= 5*8.66
= 43.3 km^2
which no questions bro.
Step-by-step explanation:
plz give me the questions
X = -b/2a = 8/2 = 4
y = 16 -8*4 -9 = 16 - 32 -9 = -25
the wertex is (4,-25)
Answer:
12 meters
Step-by-step explanation:
Draw a picture of the two pathways 30° apart. Add a circle representing the illumination of the light on one of the pathways. Draw a line from the center of the circle to the last point where it intersects the pathway without lights.
This forms a triangle. One leg is x, the length of the pathway without lights. Another leg is 5n, where n is an integer. This represents how far the light is from where the pathways meet. The third and final leg is 6, the radius of the illumination.
Use law of cosine to solve:
6² = x² + (5n)² − 2x(5n) cos 30°
36 = x² + 25n² − 5√3 xn
0 = x² − 5√3 xn + 25n² − 36
In order to have a solution, the discriminant must be greater than or equal to 0.
b² − 4ac ≥ 0
(-5√3 n)² − 4(1)(25n² − 36) ≥ 0
75n² − 100n² + 144 ≥ 0
144 − 25n² ≥ 0
144 ≥ 25n²
144/25 ≥ n²
12/5 ≥ n
So n must be an integer less than 12/5, or 2.4. Therefore, the largest value of n is 2. Substituting:
0 = x² − 5√3 x(2) + 25(2)² − 36
0 = x² − 10√3 x + 64
Solve with quadratic formula:
x = [ 10√3 ± √(300 − 4(1)(64)) ] / 2(1)
x = (10√3 ± √44) / 2
x = 5√3 ± √11
x ≈ 5.34 or 11.98
We want the larger value of x. So approximately 12 meters of the pathway without lights is illuminated.
Answer:
98%
Step-by-step explanation:
Percent means out of 100
49/50
Multiply the top and bottom by 2
49*2 = 98
------------
50*2 = 100
98/100
The percent is 98%