Answer: f(x) = -0.016x² + 1.6x
Step-by-step explanation:
If the lenght of the road over the arch is 100m, we can consider a coordinate plane and say that the road starts at point (0,0) and finishes at (100,0). The vertice of the parabola is at point (50,40), because the maximum height is 40 and it is always in the middle point of the roots.
So, we have
(0,0) (50,40) (100,0)
A quadratic function is always on the form: f(x) = ax² + bx + c
0 = a0² + b0 + c
40 = a50² + b50 + c
0 = a100² + b100 + c
0 = a0² + b0 + c → c = 0 ∴
40 = a50² + b50
0 = a100² + b100
_________________________
2500a + 50b = 40 (*2)
10000a + 100b = 0
_________________________
5000a + 100b = 80
10000a + 100b = 0 (-)
__________________________
-5000a = 80
-a=80/5000
a=-0.016
∴
2500a + 50b = 40
2500.(-0.016) + 50b = 40
-40 + 50b = 40
50b = 80
b = 80/50
b = 1.6
This way f(x) = -0.016x² + 1.6x
Answer:
5 to 2
Step-by-step explanation:
You can simplify both 20 and 8 by 4.
20/4 = 5
8/4 = 2
Answer:
the amswer is 5 hope this helps you
Answer:
4 (2 a + 3 b)
Step-by-step explanation:
Simplify the following:
8 a + 12 b
Factor 4 out of 8 a + 12 b:
Answer: 4 (2 a + 3 b)
Answer:
The area of triangle for the given coordinates is 1.5
Step-by-step explanation:
Given coordinates of triangles as
A = (0,0)
B = (3,4)
C = (3,2)
So, The measure of length AB = a = 
Or, a = 
Or, a = 
Or, a = 
∴ a = 5 unit
Similarly
The measure of length BC = b = 
Or, b = 
Or, a = 
Or, b = 
∴ b = 2 unit
And
So, The measure of length CA = c = 
Or, c = 
Or, c = 
Or, c = 
∴ c =
unit
Now, area of Triangle written as , from Heron's formula
A = 
and s = 
I.e s = 
Or. s = 
So, A = 
Or, A = 
Or, A =
× 
∴ Area of triangle = 1.5
Hence The area of triangle for the given coordinates is 1.5
Answer