After trial and error, you discover that - 2 works.
-2 || 3 16 18 -4
-6 -20 4
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3 10 - 2 0
What you have left is a quadratic
3x^2 + 10x - 2 You can just use the quadratic formula to solve for the other 2 roots.
x1 = [ - 10 +/- sqrt(10^2 - 4*3*(-2) ) ] / 6
x1 = [ -10 +/- sqrt (100 + 24) ) /6
x1 = [ - 10 +/- sqrt(124) ) / 6
x1 = [ - 10 +/- 11.14 ] / 6 = 0.1893
x2 = [ - 21.14 ] / 6 = - 3.522
Answer
x1 = - 2
x2 = 0.1893
x3 = - 3.522
A 57.6 B. 100.8 c.78.2 I’m not sure
The answer is 73,990.0 hope i help
Answer:
Domain: [-2, 0, 2, 4]
Range: [3]
Step-by-step explanation:
The domain of the function are all values of x that are plotted on the horizontal axis (x-axis), while the range are the corresponding y-values plotted on the vertical axis (y-axis).
Therefore,
Domain of the function = [-2, 0, 2, 4]
Range of the function = [3] (only 1 possible value of y can be seen as plotted on the gray]
The formula for illuminance is given by
E = I / d^2
This formula only holds true for one-dimensional illuminance
The problem asks for the illuminance across the floor. We need to use two variables, x and y.
From Pythagorean Theorem
d^2 = x^2 + y^2
and from Trigonometry
x = d cos t
y = d sin t
The function for the illuminance can be represented by the composite function
E = I cos² t / x²
and
E = I sin² t / y²
The boundary of these functions is:
<span>0 < t < 8
So, the value of t must be in radians and not in degrees</span>
