If two points of a ray lie in a plane, then all of the ray lies in the plane.true or false

1 answer:

3 0

The correct answer for the given statement above would be FALSE. If two points of a ray lie in a plane, then not all of the ray lies in the plane because i<span>f two points of a ray lie in a plane, that does not imply that the ray lies in the plane. If A then B, B does not necessarily imply A. This makes the statement incorrect. </span>

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Help me pls none of this makes sense:(

SVEN [57.7K]

Step-by-step explanation:

x+48=180

x=180-48

x=132

hope this helps you

have a nice day:)

4 0

3 years ago

Read 2 more answers

If the mountain is 1,200 feet tall, find the length of the ski lift

Nina [5.8K]

The length of the sky lift is 3.387.05 ft
Step-by-step explanation;
By an online search, i found that the sky lift
rises at an angle of 20.75 degrees.
Then we can think on a triangle rectangle, such
that:
One cathetus is 1200ft, this is the opposite
cathetus to the angle of 20.75°
The length of the ski lift would be the
hypotenuse of this triangle rectangle.
Then we can use the relationship:
Sin(a) = opposite cathetus/hypoteuse.
such that:
a = 20.75°
opposite cathetus=1200ft
then:
sin(20.75°) = 1200ft/hypotenuse
hypotenuse =1200ft/sin(20.759) =3.387.05 ft
The length of the sky lift is 3.387.05 ft

7 0

2 years ago

How would I do part A. I showed my work but I'm slightly off

ahrayia [7]

$\bf \stackrel{\textit{perimeter of the rectangle}}{P_{rect}=2x+y}~\hfill \stackrel{\textit{area of the rectangle}}{A_{rect}=xy}
\\\\\\
\stackrel{\textit{perimeter of the semi-circle, with }r=\frac{y}{2}}{P_{semic}=\cfrac{\pi y}{2}}~\hfill \stackrel{\textit{area of the semi-circle with }r=\frac{y}{2}}{A_{semic}=\cfrac{\pi y^2}{8}}
\\\\[-0.35em]
\rule{34em}{0.25pt}$

$\bf \stackrel{\textit{perimeter of the enclosure}}{60=2x+y+\cfrac{\pi y}{2}}\implies \stackrel{\textit{multiplying by 2}}{120=4x+2y+\pi y}
\\\\\\
120-2y-\pi y=4x\implies \cfrac{120-2y-\pi y}{4}=x
\\\\[-0.35em]
\rule{34em}{0.25pt}$

$\bf \stackrel{\textit{area of the enclosure}}{A=xy+\cfrac{\pi y^2}{8}}\implies A=\left( \cfrac{120-2y-\pi y}{4} \right)y+\cfrac{\pi y^2}{8}
\\\\\\
A=\cfrac{120y-2y^2-\pi y^2}{4}+\cfrac{\pi y^2}{8}\implies A=\cfrac{240y-4y^2-2\pi y^2+\pi y^2}{8}
\\\\\\
A=\cfrac{240y-4y^2-\pi y^2}{8}\implies A=\cfrac{240y}{8}-\cfrac{4y^2}{8}-\cfrac{\pi y^2}{8}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
~\hfill A=30y-\cfrac{1}{2}y^2-\cfrac{1}{8}\pi y^2~\hfill$