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
The correct answer is
the flagpole is <span>
33 feet high</span>.
Explanation:
Please refer to the attached picture.
We know:
CD = 40 feet
AC = 5 feet
∠BDC = α = 35°
Using trigonometry, we know that the definition of the tangent of an angle is the ratio between the opposite side and the adjacent side, therefore:
tan α = BC / CD
Solving for BC:
BC = CD · <span>tan α
= 40 </span>· tan (35)
= 28 feet
In order to find the height of the flagpole, we need to add the distance of the clinometer from the ground:
AB = BC + AC
= 28 + 5
= 33
Hence, the flagpole is
33 feet high.
A. After the mod. One wall is “x” longer than original (horizontal in diagram) and the other is “x” shorter (vertical)
(30+x)(30-x)
b. He original area = 30 x 30 = 900 sq. ft.
The mod. Area is (30+6)(30-6) = 36x24= 764 sq. ft.
So the room was largest to begin with when it was a square
The ratio is 1 to 68.8 or 1 : 61.8 and you're looking for 6 : x so all you have to do is multiply 61.8 by 6 and you get 370.8
