Answer:
Step-by-step explanation:
9) PQR Is an isosceles triangle
=> ∠PRQ = (180° - x)/2
PRS is an isosceles right triangle
=> ∠PRS = 45°
Have: ∠PRS + ∠PRQ = 115°
=>
=> 180 - x = (115 - 45).2 = 140
<=> x = 180 - 140 = 40
10) ABD is an isosceles right triangle => ∠ABD = 45°
BCD is an equilateral triangle => ∠CBD = 60°
have: x = ∠ABD + ∠CBD = 45° + 60° = 105°
11) have: x = y (2)
PQT is an isosceles triangle => ∠PQT = 180 - 70.2 = 40
QTS is an isosceles triangle => ∠TQS = 180 -2x
QRS is an isosceles triangle => ∠RSQ = y
have: 40 + 180 - 2x + y = 180 => 2x - y = 40 (1)
(1)(2) =>
=> x + y = 80
12) EFJ Is an equilateral triangle => ∠FJE = 60
∠FJE is the outer angle of the triangle FHJ but FHJ is an isosceles triangle
=> 60 = 2.∠JHF => ∠JHF = 30°
∠JHF is the outer angle of the triangle FHG
=> 30° = 2x
<=> x = 15°
9514 1404 393
Answer:
a (3,-1)
Step-by-step explanation:
The number that "completes the square" is the square of half the x-coefficient, (-6/2)^2 = 9. Rearranging the given function to include the square trinomial, we have ...
f(x) = x^2 -6x +9 -1 . . . . . . . here, we have 8 = 9 - 1
f(x) = (x -3)^2 -1 . . . . . . . . . . vertex form
Comparing this to the generic vertex form ...
f(x) = (x -h)^2 +k . . . . . . . vertex at (h, k)
we see that h=3 and k=-1.
The vertex is (h, k) = (3, -1).
