If Ori gives a friend three reasons for preferring soccer to basketball, that is an algorithm.
2 answers:
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Answer:
y = -1/24 x³ + 5/12 x² − 35/24 x + 25/12
Explanation:
A cubic has the form:
y = ax³ + bx² + cx + d
Given four points, we can write a system of equations:
1 = a + b + c + d
1/2 = 8a + 4b + 2c + d
1/3 = 27a + 9b + 3c + d
1/4 = 64a + 16b + 4c + d
Solving this algebraically would be time-consuming, but we can use matrices to make it easy.
First, we find the inverse of the coefficient matrix. This is messy to do by hand, so let's use a calculator:
Now we multiply by the solution matrix (again using a calculator):
So the cubic is:
y = -1/24 x³ + 5/12 x² − 35/24 x + 25/12
Answer:
the width of the turning roadway = 15 ft
Explanation:
Given that:
A ramp from an expressway with a design speed(u) = 30 mi/h connects with a local road
Using 0.08 for superelevation(e)
The minimum radius of the curve on the road can be determined by using the expression:
where;
R= radius
= coefficient of friction
From the tables of coefficient of friction for a design speed at 30 mi/h ;
= 0.20
So;
R = 214.29 ft
R ≅ 215 ft
However; given that :
The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.
From the tables of "Design widths of pavement for turning roads"
For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation
Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.
As such in Case 1 operation that falls under traffic condition B in accordance with the Design widths of pavement for turning roads;
If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft
Hence; the width of the turning roadway = 15 ft