Which of the following is the solution to the equation x + 6.8 = 53.5?
2 answers:
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Answer:
#1: Josue's dogs
#2: see images
#3: Am group median: 25.5 Pm group median: 19
#4: Am group first Quartile = 16 Pm group first Quartile = 10
#5: Am group third Quartile = 30 Pm group third Quartile = 37
#6: see images
#7: Am group Interquartile Range = 14 Pm group Interquartile Range = 27
#8: Am group because its Interquartile gruop is smaller than the Pm group
Step-by-step explanation:
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Let the point_1 = p₁ = (1,4)
and point_2 = p₂ = (-2,1)
and Point_3 = p₃ = (x,y)
The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)
The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d =
d₁ =
d₂ =
d₁ = d₂
∴
⇒⇒ eliminating the root
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
(x-1)²+(y-4)² = 18
from equatoin (1) y-4 = 1-x
∴(x-1)²+(1-x)² = 18 ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 =
∴ x = 4 or x = -2
∴ y = 1 or y = 7
Point_3 = (4,1) or (-2,7)
and point_2 = p₂ = (-2,1)
and Point_3 = p₃ = (x,y)
The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)
The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d =
d₁ =
d₂ =
d₁ = d₂
∴
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
(x-1)²+(y-4)² = 18
from equatoin (1) y-4 = 1-x
∴(x-1)²+(1-x)² = 18 ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 =
∴ x = 4 or x = -2
∴ y = 1 or y = 7
Point_3 = (4,1) or (-2,7)