Find two consecutive odd integers whose sum is 36. Which of the following equations could be used to solve the problem?

1 answer:

5 0

Let's say your equation is 2x+2=36
if you solve the equation you get 17=x then you do 36-17 and you get 19

so your answers are 17 and 19

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What is m in the simultaneous equation <br> M -n=1<br><br> M+n=3

Lady bird [3.3K]

The simultaneous equations is missing +8

8 0

2 years ago

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Calculate pt3 such that a line from pt1 to pt3 is perpendicular to the line from pt1 to pt2, and the distance between pt1 and pt

Leni [432]

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 = $\sqrt{Δx^2+Δy^2}$
d₁ = $\sqrt{(-2-1)^2+(1-4)^2}$
d₂ = $\sqrt{(x-1)^2+(y-4)^2}$
d₁ = d₂
∴ $\sqrt{(-2-1)^2+(1-4)^2} = \sqrt{(x-1)^2+(y-4)^2}$ ⇒⇒ 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 = $\pm \sqrt{9} = \pm 3$
∴ x = 4 or x = -2
∴ y = 1 or y = 7

Point_3 = (4,1) or (-2,7)

8 0

3 years ago

5x + 6(x - 2) - 8 (x - 3)

Alex Ar [27]

Answer:

The value of $x=-4$ . The figure is also attached below.