SOMEONE PLEASE HELP. SHOW WORK

Mathematics

1 answer:

kirill [66]3 years ago

8 0

You need to break this into smaller shapes. This shape consists of 4 smaller shapes: 3 triangles and 1 rectangle.

Formulas:
Triangle: A = 1/2 bh
Rectangle: A= bh

Let's start with the triangle on top with the points B, C and D.
Base = 6
Height = 3

A = 1/2 (6)(3)
Multiply base by height
A = 1/2 (18)
Multiply.
A= 9

The triangle on the left:
Base= 4
Height = 4

A= 1/2 (4)(4)
A= 1/2 (16)
A= 8

The triangle on the right:
Base= 2
Height= 4

A= 1/2 (2)(4)
A= 1/2 (8)
A= 4

The rectangle in the middle:
Base= 6
Height=4

A= (6)(4)
A= 24

Now add all the areas together:
9 + 8 + 4 + 24 = 45 cm²

Hope this helps! :)

You might be interested in

Let C be the positively oriented square with vertices (0,0), (1,0), (1,1), (0,1). Use Green's Theorem to evaluate the line integ

liq [111]

Answer:

1/2

Step-by-step explanation:

The interior of the square is the region D = { (x,y) : 0 ≤ x,y ≤1 }. We call L(x,y) = 7y²x, M(x,y) = 8x²y. Since C is positively oriented, Green Theorem states that

$\int\limits_C {L(x,y)} \, dx + {M(x,y)} \, dy = \int\limits^1_0\int\limits^1_0 {(Mx - Ly)} \, dxdy$

Lets calculate the partial derivates of M and L, Mx and Ly. They can be computed by taking the derivate of the respective value, treating the other variable as a constant.

Mx(x,y) = d/dx 8x²y = 16xy
Ly(x,y) = d/dy 7y²x = 14xy

Thus, Mx(x,y) - Ly(x,y) = 2xy, and therefore, the line ntegral is equal to the double integral

$\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy$

We can compute the double integral by applying the Barrow's Rule, a primitive of 2xy under the variable x is x²y, thus the double integral can be computed as follows

$\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy = \int\limits^1_0 {x^2y} |^1_0 \,dy = \int\limits^1_0 {y} \, dy = \frac{y^2}{2} \, |^1_0 = 1/2$