Find the area of the region bounded by the line y=3x−6 and line y=−2x+8. a) the x-axis.
1 answer:
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We have that
csc ∅=13/12
sec ∅=-13/5
cot ∅=-5/12
we know that
csc ∅=1/sin ∅
sin ∅=1/ csc ∅------> sin ∅=12/13
sec ∅=-13/5
sec ∅=1/cos ∅
cos ∅=1/sec ∅------> cos ∅=-5/13
sin ∅ is positive and cos ∅ is negative
so
∅ belong to the II quadrant
therefore
<span>the coordinates of point (x,y) on the terminal ray of angle theta are
</span>x=-5
y=12
the answer is
point (-5,12)
see the attached figure
csc ∅=13/12
sec ∅=-13/5
cot ∅=-5/12
we know that
csc ∅=1/sin ∅
sin ∅=1/ csc ∅------> sin ∅=12/13
sec ∅=-13/5
sec ∅=1/cos ∅
cos ∅=1/sec ∅------> cos ∅=-5/13
sin ∅ is positive and cos ∅ is negative
so
∅ belong to the II quadrant
therefore
<span>the coordinates of point (x,y) on the terminal ray of angle theta are
</span>x=-5
y=12
the answer is
point (-5,12)
see the attached figure
Answer:
64.3 lumens
Step-by-step explanation:
Data provided in the question:
Length of the testing chamber = 35 cm
Light at one end = 10 lumens
Light at other end = 200 lumens
now,
since the variation is linear,
therefore, the gradient of the variation
=
= 5.43
also,
the darkest end is the end with 10 lumens
therefore,
The lumens at 10 cm from the darkest end
= 10 + 5.43(10)
= 10 + 54.3
= 64.3 lumens