Respuesta :

lublana

Answer:

[tex]AC=8x-1[/tex]

[tex]QR=5y+20[/tex]

Step-by-step explanation:

We are given that

[tex]AB=x+2[/tex]

[tex]BC=7x-3[/tex]

[tex]PQ=8y+5[/tex]

[tex]PR=13y+25[/tex]

We have to find the value of AC and QR

[tex]AC=AB+BC[/tex] (segment addition property)

Substitute the value

Then, we get

[tex]AC=x+2+7x-3=8x-1[/tex]

[tex]PQ+QR=PR[/tex]  (segment addition property)

Substitute the values then we get

[tex]8y+5+QR=13y+25[/tex]

[tex]QR=13y+25-8y-5[/tex]

Using subtraction property of equality

[tex]QR=5y+20[/tex]

fichoh

The simplified expressions for the length of the line segment are :

  • AC = 8x - 1
  • QR = 5y + 20

  • The line segment AC equals:

AC = AB + BC ;

AB = x+2 ; BC = 7x-3

AC = x + 2 + 7x - 3

AC = 8x - 1

  • Line segment PR

PR = PQ + QR - - (1)

PR = 13y + 25 ; PQ = 8y + 5

From (1)

QR = PR - PQ

Hence,

QR = 13y + 25 - (8y + 5)

QR = 13y + 25 - 8y - 5

QR = 5y + 20

Therefore, the line segments are :

  • AC = 8x - 1
  • QR = 5y + 20

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