The fraction 2/9 is equal to 2 ÷ 9. Use long division to find the decimal value of 2/9. Show your work.
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Answers:
Vertical asymptote: x = 0
Horizontal asymptote: None
Slant asymptote: (1/3)x - 4
<u>Explanation:</u>
d(x) =
=
Discontinuities: (terms that cancel out from numerator and denominator):
Nothing cancels so there are NO discontinuities.
Vertical asymptote (denominator cannot equal zero):
3x ≠ 0
<u>÷3</u> <u>÷3 </u>
x ≠ 0
So asymptote is to be drawn at x = 0
Horizontal asymptote (evaluate degree of numerator and denominator):
degree of numerator (2) > degree of denominator (1)
so there is NO horizontal asymptote but slant (oblique) must be calculated.
Slant (Oblique) Asymptote (divide numerator by denominator):
- <u>(1/3)x - 4 </u>
- 3x) x² - 12x + 20
- <u>x² </u>
- -12x
- <u>-12x </u>
- 20 (stop! because there is no "x")
So, slant asymptote is to be drawn at (1/3)x - 4
The depth of the swimming pool that is filled to the top is; 4 m
<h3>Snell's Law</h3>
I have attached a schematic diagram showing this question.
The correct width of the pool is 4 meters. Thus; w = 4 m
Incident Angle; θ₁ = 20°
A right angle is 90° and so the angle θ₂ is calculated from;
θ₂ = 90° - θ₁
θ₂ = 90° - 20°
θ₂ = 70°
We can use snell's law formula to find θ₃.
Thus;
n₁sinθ₂ = n₂sinθ₃
where;
n₁ is refractive index of air = 1
n₂ is refractive index of water = 1.33
Thus;
1*sin 70 = 1.33*sin θ₃
sin θ₃ = (sin 70)/1.33
Solving this gives;
θ₃ = 44.95°
By usage of trigonometric ratios we can find the depth of the pool using;
w/d = tan θ₃
Thus;
d = w/(tan θ₃)
d = 4/(tan 44.95)
d ≈ 4 m
Read more about Snell's Law at; brainly.com/question/10112549
