The depth of the water at the <em>deepest</em> point in the waterslide and to the nearest hundredth of a meter is approximately 0.19 meters. (Correct choice: A)
<h3>How to determine the depth of the water in a waterslide</h3>
In this question we should apply the concepts of <em>right</em> triangles and <em>trigonometric</em> functions to determine the height of the water within the waterslide. A geometric diagram of the <em>cross</em> section of the waterslide is presented below, which indicates the existence of <em>circular</em> symmetry.
Now we proceed to determine the height of the water:
cos α = (0.5 m/0.75 m)
α ≈ 48.190°
y = (0.75 m) · sin 48.190°
y = 0.559 m
x = 0.75 m - 0.559 m
x = 0.191 m
The depth of the water at the <em>deepest</em> point in the waterslide and to the nearest hundredth of a meter is approximately 0.19 meters. (Correct choice: A)
To learn more on trigonometry: brainly.com/question/22698523
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In a parallelogram, the diagonals bisect each other.
Therefore, x = 15.
Answer:
Step-by-step explanation:
V=hpir^2
so
A:B=
4pi1^2:4pi2^2=
4pi1:4pi4=
4pi:16pi=
4:16=
1:4
the ratio of volume of A to B is 1:4