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3 years ago

7

2 lines intersect. A line with points A, B, D intersects a line with points E, B, C at point B, forming 4 angles. In the diagram

, which angles are vertical angles? Select two options. AngleABE and AngleABC AngleABE and AngleCBD AngleABE and AngleEBD AngleABC and AngleEBD AngleABC and AngleCBD

1 answer:

solmaris [256]3 years ago

7 0

Answer:

Its B and D

Step-by-step explanation:

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What happens to energy during a change of state? Why is it a physical change?

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Answer & Explanation:

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3 years ago

Find the sum of the exterior angles of a 41-gon.<br> *

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which of the following side lengths can form a triangle. A.10,16,&30 B.11,14,&26. C.41,51,&61. D.100,200,&600.

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3 years ago

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

Solve the equation for all real solutions in simplest form.<br> z^2 - 12z +9= -3

Mars2501 [29]

Answer:

Solving the equation for all real solutions in simplest form.

$z^2 - 12z +9= -3$ we get $\mathbf{z=10.9\: or\: z=1.1}$

Step-by-step explanation:

We need to solve the equation for all real solutions in simplest form.

$z^2 - 12z +9= -3$

First simplifying the equation:

$z^2 - 12z +9+3= -3+3\\z^2 - 12z +12= 0$

Now, we can solve the equation using quadratic formula:

$z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

we have a = 1, b=-12 and c=12

Putting values in formula and finding values of x

$z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\z=\frac{-(-12)\pm\sqrt{(-12)^2-4(1)(12)}}{2(1)}\\z=\frac{12\pm\sqrt{144-48}}{2}\\z=\frac{12\pm\sqrt{96}}{2}\\z=\frac{12\pm9.8}{2}\\z=\frac{12+9.8}{2}\:or\:z=\frac{12-9.8}{2}\\z=10.9\:or\:z=1.1$

So, we get value of z: z=10.9 or z=1.1

Solving the equation for all real solutions in simplest form.

$z^2 - 12z +9= -3$ we get $\mathbf{z=10.9\: or\: z=1.1}$

3 0

3 years ago