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yuradex [85]
1 year ago
5

If Angle 6 is congruent to angle 10 and Angle 5 is congruent to angle 7, which describes all the lines that must be parallel? Li

nes r and s are crossed by lines t and u to form 16 angles. Clockwise from top left, at the intersection of r and t, the angles are 1, 2, 3, 4; at the intersection of s and t, 5, 6, 7, 8; at the intersection of u and s, 9, 10, 11, 12; at the intersection of u and r, 13, 14, 15, 16. Only lines r and s must be parallel. Only lines t and u must be parallel. Lines r and s and lines t and u must be parallel. Neither lines r and s nor lines t and u must be parallel.

Mathematics
2 answers:
mel-nik [20]1 year ago
8 0

Answer:

quick answer is the Third option-  Lines "r" and "s" and lines "t" and "u" must be parallel.

Nina [5.8K]1 year ago
7 0

Answer:

Third option: Lines "r" and "s" and lines "t" and "u" must be parallel.

Step-by-step explanation:

The missing figure is attached.

You need to remember that:

1- A Transversal is defined as a line that intersects two or more lines.

2- When a transversal cut two parallel lines, several angles are formed, which are grouped in pairs. Some of them are:

a. Vertical angles: are those pairs of angles that share the same vertex and are opposite to each other. These angles are congruent.

b. Corresponding angles: are those  pairs of non-adjacent angles located on the same side of the transversal and outside the parallel lines. They are congruent.

In this case, you can identify in the figure that:

 \angle 6 and \angle 10 are Corresponding angles.

\angle5 and \angle 7 are Vertical angles.

Therefore, based on the explained before, you can conclude that lines "r" and "s" and lines "t" and "u", must be parallel.

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In this rectangular box, EF = 16, FD = 5, and DB = 30. Find AF.
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Answer:

FA = 25.87  

Step-by-step explanation:

Given: BD = 30 , DF = 5 , EF = 16

To find: FA

Rectangular box is a Cuboid.

Figure attached.

AB = FE = 16

So,

In  ΔADB

using Pythagoras theorem,

DB^2=AB^2+AD^2\\30^2=16^2+AD^2\\900=256+AD^2\\AD^2=900-256\\AD^2=644\\AD=\sqrt{644}\\AD=2\sqrt{161}

Again using Pythagoras theorem in Δ ADF we get

FA^2=DF^2+AD^2\\FA^2=5^2+(2\sqrt{161})^2\\FA^2=25+644\\FA^2=669\\FA=\sqrt{669}\\FA=25.87

Therefore, FA = 25.87

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A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. Find the lengths of the median o
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Answer:

5/2units\\.

\sqrt{47/2}units \\.

\sqrt{41/2}units\\.

Step-by-step explanation:

The given points are A(1,2,3), B(-2,0,5) and C(4,1,5). The triangle is represented in the attach file where the three possible median are length AE, BF, and CD. We determine the coordinate of point D,E and F using the midpoint equation which is for any point A(x,y,z) and point B(a,b,c), the midpoint D is determine by

D=(\frac{x+a}{2},\frac{y+b}{2},\frac{z+c}{2})\\.

Hence going by the above formula we determine the coordinate of point D,E and F

D=(\frac{-2+1}{2},\frac{0+2}{2},\frac{5+3}{2})\\.

D=(\frac{-1}{2},1,4)\\.

point E

E=(\frac{4-2}{2},\frac{1+0}{2},\frac{5+5}{2})\\.

E=(1,\frac{1}{2},5)\\.

Point F

F=(\frac{4+1}{2},\frac{1+2}{2},\frac{5+3}{2})\\.

F=(\frac{5}{2},\frac{3}{2},4)\\.

To determine the length of each median line we use the formula for distance between two points which is express as

AB=\sqrt{(y_{2}-y_{1} )^{2} +(x_{2}-x_{1} )^{2}+(z_{2}-z_{1} )^{2}} \\.

Using the above formula we determine the length of line AE,BF and CD.

AE=\sqrt{(1-1 )^{2} +(1/2-2)^{2}+(5-3 )^{2}} \\.

AE=\sqrt{0 +9/4+4} \\.

AE=\sqrt{25/4} \\.

AE=5/2units\\.

For point BF

BF=\sqrt{(5/2+2 )^{2} +(3/2-0)^{2}+(4-5 )^{2}}\\.

BF=\sqrt{81/4 +9/4+1} \\.

BF=\sqrt{47/2} \\.

BF=\sqrt{47/2}units \\.

For point CD

CD=\sqrt{(-1/2-4 )^{2} +(1-1)^{2}+(4-5 )^{2}}\\.

BF=\sqrt{81/4 +0+1} \\.

BF=\sqrt{41/2} \\.

BF=\sqrt{41/2}units\\.

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