Please help asap!

Mathematics

1 answer:

lidiya [134]2 years ago

8 0

You would distribute, 300 times 80 is 2400 and -3x times 1x is -3x so the equation would be y=2400-3x

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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

You are standing above the point (5,5)(5,5) on the surface z=15−(2x2+2y2)z=15−(2x2+2y2). (a) in which direction should you walk

Marianna [84]

Since the surface is the top part of a sphere centred on the origin, the directions of steepest descent are any vector originating from the origin (0,0).

At (5,5), the direction would be dictated by the line joining (5,5) and the origin (0,0), i.e. along <5,5> or equivalently along <1,1>.

You can also go through vector calculus to arrive at the same result.