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The value of k such that the graph of f and the graph of g only intersect one is equal to 2.
The value of k such that the graph of f and the graph of g only intersect one is equal to 2. According to the image attached below, functions f(x) and g(x) intersect at point (x, y) = (0, 4) for k = 2.
<h3>How to find the value of the constant k of a system of two polynomic equations</h3>
Herein we have a system formed by two <em>nonlinear</em> equations, a <em>quadratic</em> equation and a <em>cubic</em> equation. Given the constraint that both function must only intersect once, we have the following expression:
f(x) - x² - k · x = 4 (1)
g(x) - x² - k · x = x³ + 2 · k (2)
x³ + 2 · k = 4
x³ + 2 · (k - 2) = 0
If f and g must intersect once, then the roots must of the form:
(x - r)³ = x³ + 2 · (k - 2)
x³ - 3 · r · x² + 3 · r² · x - r³ = x³ + 2 · (k - 2)
Then, the following conditions must be met: - 3 · r · x² = 0, 3 · r² · x = 0. If x may be any real number, then r must be zero and the value of k must be:
2 · (k - 2) = 0
k - 2 = 0
k = 2
Therefore, the value of k such that the graph of f and the graph of g only intersect one is equal to 2. According to the image attached below, functions f(x) and g(x) intersect at point (x, y) = (0, 4) for k = 2.
To learn more on polynomic functions: brainly.com/question/24252137
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Answer:
Step-by-step explanation:
The area of a circular sector with a central angle θ and radius r is given by:
<em>Note</em>: The angle must be in radians.
We need to find both the radius and the angle in radians.
Diameter = 9.6 cm
Radius r=Diameter/2=4.8 cm
Angle in radians= Angle in degrees*π / 180
θ = 140*π / 180 = 2.4435 rad
Now we calculate the area: