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>
The Answer Is A)11
Explanation: Subtract 3 From Both Sides By 3 Then Divide By 3
Like XZ divides the cord YV into two congruent parts (YW=5.27 cm=WV), this segment XZ must be perpendicular to the segment YV, then the angle XWY in triangle XWY is a right angle (90°) and the triangle XWY is a right angle.
We can apply the trigonometric ratios in triangle XWY:
Hypotenure: XY
sin 44°=(Opposite leg to 44°)/(hypothenuse)
sin 44°=YW/XY
sin 44°=(5.27 cm)/XY
Solving for XY. Cross multiplication:
sin44° XY=5.27 cm
Dividing both sides of the equation by sin 44°:
sin 44° XY / sin 44° = (5.27 cm)/sin 44°
XY=(5.27/sin 44°) cm
XY=(5.27/0.694658370) cm
XY=7.586462929 cm
This value XY is the radius of the circle, then:
XZ=XY→XZ=7.586462969 cm
tan 44°=(Opposite leg to 44°) / (Adjacent leg to 44°)
tan 44°=YW/XW
tan 44°=(5.27 cm)/XW
Solving for XW. Cross multiplication:
tan 44° XW=5.27 cm
Dividing both sides of the equation by tan 44°:
tan 44° XW / tan 44°=(5.27 cm)/tan 44°
XW=(5.27/tan 44°) cm
XW=(5.27/0.965688775) cm
XW=5.457244753 cm
WZ=XZ-XW
WZ=7.586462969 cm-5.457244753 cm
WZ=2.129218216 cm
Rounded to 2 decimal places:
WZ=2.13 cm
Answer: The <span>measurement is closest to the measure of segment WZ is
2.13 cm</span>
Answer:graph B
Step-by-step explanation:because if a line is parallel they have no solutions with a certain slope
If two lines intersect then they must have at least a solution
