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elixir [45]
3 years ago
14

ASAP!!!!!!!!! PLEASE help me with this question! This is really urgent! No nonsense answers please.

Mathematics
1 answer:
Cloud [144]3 years ago
7 0
<h3>Answer: C) increase from 45 degrees to 50 degrees</h3>

===============================================

Explanation:

Let's calculate the angle B based off the arcs CDF and GHJ

B = (far arc - near arc)/2

B = (arc CDF - arc GHJ)/2

B = (130 - 40)/2

B = 90/2

B = 45

----------

Now let's change GHJ to 30 degrees, while keeping the other arc the same.

B = (far arc - near arc)/2

B = (arc CDF - arc GHJ)/2

B = (130 - 30)/2

B = 100/2

B = 50

Angle B has increased from 45 degrees to 50 degrees.

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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
maks197457 [2]

Answer:

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

3 0
3 years ago
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