3 years ago

Pleaseeeee helppppp lol

Mathematics

2 answers:

Molodets [167]3 years ago

8 0

Answer:

2/4 I believe

Step-by-step explanation:

trapecia [35]3 years ago

6 0

2/4

Step-by-step explanation:

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An elliptical culvert is 3.5 feet tall and 6 feet wide. It is filled with water to a depth of 0.95 feet. Find the width of the s

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$W_s \approx5.3$

Step-by-step explanation:

From the question we are told that

Height $H=3.5ft$

Weight $W= 6ft$

Depth $D=0.95ft$

Generally the equation for ellipses is given as

$\frac{x^2}{a^2} +\frac{y^2}{b^2} =1$

Where

$b=H/2\\b=3.5/2\\Therefore\\b=1.75\\a=W/2\\a=6/2\\Therefore\\a=3\\$

$\frac{x^2}{3^2} +\frac{y^2}{1.75^2} =1$

Generally to find y in the equation $\frac{x^2}{a^2} +\frac{y^2}{b^2} =1$

$y=-b+d$

$y=-1.75+0.95$

$y=-0.8$

Therefore

$\frac{x^2}{3^2} +\frac{(0.8)^2}{1.75^2} =1$

$\frac{x^2}{3^2} =1-\frac{(0.8)^2}{1.75^2}$

$\frac{x^2}{3^2} =1-0.2089795918$

$X^2 =3^2(1-0.2089795918)$

$X^2 =7.119183674$

$X =\sqrt{7.119183674}$

$X =2.66817984$

Therefore the width of the given stream is

$W_s=2.66817984*2$

$W_s=5.33635968$

$W_s \approx5.3$

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