To find where on the hill the canonball lands
So 0.15x = 2 + 0.12x - 0.002x^2
Taking the LHS expression to the right and rearranging we have -0.002x^2 + 0.12x -
0.15x + 2 = 0.
So we have -0.002x^2 - 0.03x + 2 = 0
I'll multiply through by -1 so we have
0.002x^2 + 0.03x -2 = 0.
This is a quadratic eqn with two solutions x1 = 25 and x2 = -40 since x cannot be negative x = 25. The second solution y = 0.15 * 25 = 3.75
Answer:
The values of x for which the model is 0 ≤ x ≤ 3
Step-by-step explanation:
The given function for the volume of the shipping box is given as follows;
V = 2·x³ - 19·x² + 39·x
The function will make sense when V ≥ 0, which is given as follows
When V = 0, x = 0
Which gives;
0 = 2·x³ - 19·x² + 39·x
0 = 2·x² - 19·x + 39
0 = x² - 9.5·x + 19.5
From an hint obtained by plotting the function, we have;
0 = (x - 3)·(x - 6.5)
We check for the local maximum as follows;
dV/dx = d(2·x³ - 19·x² + 39·x)/dx = 0
6·x² - 38·x + 39 = 0
x² - 19/3·x + 6.5 = 0
x = (19/3 ±√((19/3)² - 4 × 1 × 6.5))/2
∴ x = 1.288, or 5.045
At x = 1.288, we have;
V = 2·1.288³ - 19·1.288² + 39·1.288 ≈ 22.99
V ≈ 22.99 in.³
When x = 5.045, we have;
V = 2·5.045³ - 19·5.045² + 39·5.045≈ -30.023
Therefore;
V > 0 for 0 < x < 3 and V < 0 for 3 < x < 6.5
The values of x for which the model makes sense and V ≥ 0 is 0 ≤ x ≤ 3.
Answer:
1
Step-by-step explanation:
|3 + 2 - 6| = |5 - 6| = |-1| = 1
I’m sorry I don’t know but I need the points:(
Answer:
the point r would be located at (4,3)
Step-by-step explanation:
granted that the quadralaterel is a square that would mean that all side lengths are equal and the distance between points x and q is 3 meaning that the same would have to be said about the distance between s and r
