Answer:
3/4
Step-by-step explanation:
Use slope formula y2-y1 over x2-x1
9-3=6
5--3 becomes 5+3 which equals 8
so 6/8
which when simplified is 3/4.
Answer:
The surface area of the pyramid is 466,137 squared cubits
Step-by-step explanation:
This is simply asking us to find the surface area of the square based pyramid with the given dimensions.
The first thing we need to know is the principle for finding the surface area of such pyramidal shapes. To get the surface area of a pyramid, we will have to add the base area to the area of the side faces ( lateral area)
The base area of the square based pyramid, will be the same as the area of a square which is = 453 cubits X 453 cubits =
The lateral area has already been given to us as 260,928 squared cubits.
The surface area of the pyramid is 260,928 + 205209 = 466137 squared cubits
Hence, the surface area of the pyramid is 466137 squared cubits
Answer:
I Believe it is the first two
Step-by-step explanation:
Answer:
-6re−r [sin(6θ) - cos(6θ)]
Step-by-step explanation:
the Jacobian is ∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ
x = e−r sin(6θ), y = er cos(6θ)
δx/δθ = -6rcos(6θ)e−r sin(6θ), δx/δr = -sin(6θ)e−r sin(6θ)
δy/δθ = -6rsin(6θ)er cos(6θ), δy/δr = cos(6θ)er cos(6θ)
∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ
= -6rcos(6θ)e−r sin(6θ) × cos(6θ)er cos(6θ) - [-sin(6θ)e−r sin(6θ) × -6rsin(6θ)er cos(6θ)]
= -6rcos²(6θ)e−r (sin(6θ) - cos(6θ)) - 6rsin²(6θ)e−r (sin(6θ) - cos(6θ))
= -6re−r (sin(6θ) - cos(6θ)) [cos²(6θ) + sin²(6θ)]
= -6re−r [sin(6θ) - cos(6θ)] since [cos²(6θ) + sin²(6θ)] = 1
Answer:
(2,1)
Step-by-step explanation:
This is because when you type the equations into desmos, this is the cordinate that they collide.
Hope this answers your question.