According to the direct inspection, we conclude that the best approximation of the two solutions to the system of <em>quadratic</em> equations are (x₁, y₁) = (- 1, 0) and (x₂, y₂) = (1, 2.5). (Correct choice: C)
<h3>What is the solution of a nonlinear system formed by two quadratic equations?</h3>
Herein we have two parabolae, that is, polynomials of the form a · x² + b · x + c, that pass through each other twice according to the image attached to this question. We need to estimate the location of the points by visual inspection on the <em>Cartesian</em> plane.
According to the direct inspection, we conclude that the best approximation of the two solutions, that is, the point where the two parabolae intercepts each other, to the system of two <em>quadratic</em> equations are (x₁, y₁) = (- 1, 0) and (x₂, y₂) = (1, 2.5). (Correct choice: C)
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F ( x ) = k * x²
f ( 4 ) = 96
96 = k * 4²
96 = 16 k
k = 96 : 16
k = 6
f ( 2 ) = 6 * 2² = 6 * 4 = 24
Answer: D ) 24
X = 2/3 makes the equation true.
Answer: x = 2/3
Answer:
Step-by-step explanation:
6
Take the higher number and subtract the next lower value. It will tell you the difference.
17-11=6
Perimeter:
The perimeter of the triangle is the sum of its sides.
We have then:
P = 8 + 12 + 10
P = 30 units
Semi-perimeter:
In geometry, the semiperimeter of a polygon is half its perimeter.
s = P / 2
s = 30/2
s = 15 units.
Area:
Knowing the semiperimeter and the sides, the area is:
A = root (s * (s-a) * (s-b) * (s-c))
where,
s: semi-meter
a, b, c are sides.
A = root (15 * (15-8) * (15-12) * (15-10))
A = 39.68626967
A = 40 units ^ 2
