Given:
Compound shape
To find:
The area of the compound shape.
Solution:
The compound shape is splitted into two parallelograms.
<u>Bottom parallelogram:</u>
Base = 7.5 cm
Height = 5 cm
Area of the parallelogram = base × height
= 7.5 × 5
= 37.5 cm²
The area of the Bottom parallelogram 37.5 cm².
<u>Top parallelogram:</u>
Base = 7.5 cm
Height = 4.5 cm
Area of the parallelogram = base × height
= 7.5 × 4.5
= 33.75 cm²
The area of the top parallelogram 33.75 cm².
Compound shape = 37.5 + 33.75
= 71.25 cm²
The area of the compound shape is 71.25 cm².
Answer:
Distribute 2 to (x + 6) and 3 to (x-4)
Step-by-step explanation:
Answer:
x=1
Step-by-step explanation:
If V is the midpoint of the line UW, that would mean that there's an equal distance between UV and 9x. So, the value of UV and VW would eed to be the same.
x+8=9x
8=8x
x=1
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
10x+32=0
10x=-32
x=-32/10 or -3.2 or -16/5
