Answers:
- Plane EFGH
- Angle 2 and angle 3
- Alternate exterior angles
- See the explanation below
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Explanation:
Problem 1
The plane ABCD is the floor of the box or room.
The ceiling plane EFGH is parallel to the floor.
Parallel planes never intersect, must like how parallel lines in 2D never intersect. As such, parallel planes are the same distance apart.
In contrast, something like plane ABFE intersects with ABCD along the segment AB. This shows ABFE is not parallel to ABCD.
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Problem 2
Same side interior angles, aka consecutive interior angles, are inside the parallel (or nearly parallel) lines. One such example is angle 2 and angle 3. The other pair being angle 6 and angle 7. They must be on the same side of the transversal.
In contrast, alternate interior angles would be something like the pair angle 2 and angle 7. The other pair being angle 3 and angle 6. This time the angles are on alternating or opposite sides of the transversal.
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Problem 3
Angles 1 and 8 are exterior of the parallel (or nearly parallel) lines. They are on alternating sides of the transversal. Therefore, we consider them to be alternate exterior angles.
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Problem 4
The use of "alternate" refers to the idea the angles are on opposite or alternating sides of the transversal line. The transversal line is the line that crosses the two other lines that are almost parallel.
I like to think of the parallel (or nearly parallel) lines as train tracks. The transversal is the roadway that crosses both train tracks. Then we could have locations 1 and 8 on opposite sides of the roadways to represent the alternate exterior angles.
As you can probably guess or know by now, the "exterior" is from the angles being outside or exterior of the parallel (or nearly parallel) lines.
I use the substitution method, but you can also use elimination and matrix
1. Solve for y in 3x+y=14
y=14-3x
2. Substitute y=14-3x into x+2y=3
-5x+28=3
3. Solve for x in -5x+28=3
x=5
4. Substitute x=5 into y=14-3x
y=-1
5. Therefore,
x=5
y=-1
Have a nice day :D
See the attached picture to better understand the problem
we know that
<span>the area of the shaded portion in the circle=area 1+area 2
step 1find area 1area 1=area of semi circle
area 1=pi*r</span>²/2
diameter=12 units
radius r=12/2----> 6 units
r=6 units
area 1=pi*6²/2----> 56.52 units²
step 2
find the area 2
area 2=area sector CADC-area triangle ACD
in the right triangle ABC
BC=3
AC=r-----> AC=6
AB=?
applying Pythagoras Theorem
AC²=AB²+BC²------> AB²=6²-3²-----> AB²=27----> AB=3√3 units
area triangle ACD=b*h/2
b=2*3√3---> 6√3 units
h=3 units
area triangle ACD=6√3*3/2----> 9√3 units²-----> 15.59 units²
find the angle ACB
tan ∠ACB=AB/BC-----> 3√3/3---> √3
∠ACB=arc tan (√3)-----> 60°
the central angle ACD=60*2-----> 120°
area of sector CADC=(120/360)*pi*r²----> (120/360)*pi*6²---> 37.68 units²
area 2=area sector CADC-area triangle ACD
area 2=37.68-15.59----> 22.09 units²
step 3
the area of the shaded portion in the circle=area 1+area 2
the area of the shaded portion in the circle= 56.52 +22.09----> 78.61 units²
the answer isthe area of the shaded portion is 78.61 units²
Answer:
x=4
Step-by-step explanation: