Answer:
The perimeter of triangle PQR is 17 ft
Step-by-step explanation:
Consider the triangles PQR and STU
1. PQ ≅ ST = 4 ft (Given)
2. ∠PQR ≅ ∠STU (Given)
3. QR ≅ TU = 6 ft (Given)
Therefore, the two triangles are congruent by SAS postulate.
Now, from CPCTE, PR = SU. Therefore,

Now, side PR is given by plugging in 3 for 'y'.
PR = 3(3) - 2 = 9 - 2 = 7 ft
Now, perimeter of a triangle PQR is the sum of all of its sides.
Therefore, Perimeter = PQ + QR + PR
= (4 + 6 + 7) ft
= 17 ft
Hence, the perimeter of triangle PQR is 17 ft.
Substitute the value of x in equation 2. Then you'll get y=3. Then again substitute the value of y in x=1+y, you'll get x=4
It's False; an octagon has 8 vertices. When you remove the starting vertex and the two adjacent vertices we're left with 5 possible diagonals
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Step-by-step explanation:
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