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2 years ago

What method can you use to find the area of the composite figure? Check all that apply.

Mathematics

2 answers:

olga2289 [7]2 years ago

7 0

Answer:

Decompose the figure into two rectangles and add the areas.

Step-by-step explanation:

We can draw a vertical line segment from the smaller section of the rectangle straight down. This would cut the rectangle into 2 smaller rectangles. We can then find the area of each rectangle and add them to find the area of the composite figure.

klasskru [66]2 years ago

4 0

Answer:

Decompose the figure into two rectangles and add the areas.

Find the area of the entire rectangle and of the removed corner and subtract the areas.

Decompose the figure into three rectangles and add the areas.

Step-by-step explanation:

With all of these you can actually calculate the area of the composite figure, some of them are more easy and efficient than the other, for example dividing the composite area into three rectangles is not very efficient but will do the job, and the one where you decompose the area into two rectangles would be the best one, as well as the one where you find the area of the larger rectangle and the substract from that the rectangle that is taken off in the right corner.

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Step-by-step explanation:

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An isosceles triangle has two identical angles.

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3 years ago

A particle moves along line segments from the origin to the points (2, 0, 0), (2, 5, 1), (0, 5, 1), and back to the origin under

kondaur [170]

A particle moves along line segments from the origin to the points, work done is mathematically given as

W=184.5units

<h3>What is the solution of an equation that matches the model.?</h3>

Considering line segments from (0,0,0)origin to (2,0,0) , (2,5,1) and (0,5,1) under the infulence of force

Generally, the equation for force is mathematically given as

F = z2i + 5xyj + 2y2k

Therefore, Considering u*v

$u * v = (u_1j+u_2j+u_3k) * (v_1i+v_2j+v_3k)$

$u* v = u_1v_1(i * i) + u_1v_2(i * j)+u_1v_3(i * k) + u_2v_1(j * i) + u_2v_2(j * j)+u_2_3(j* k) + u_3v_1(k * i) + u_3v_2(k * j)+u_3v_3(k * k)$

Where

$i * i = j *j=k * k=0$

Hence

$u* v = u_1v_2k-u_1v_3j-u_2v_1k+u_2v_3i +u_3v_1j - u_3v_2i$

$u = (u_1,u_2,u_3) = (0,5,1)\\\\v = (v_1,v_2,v_3) = (-3,0,0)$

The normal equation formed

-2y + 15z = 0

z= (1/5)y

Considering the level surface and differential surface area

h(x,y,z) = -y + 5z =0

$dS = |grad(h)| dA$

In terms of the x and y coordinates of (2,0,0) and (2,5,1) and (0,5,1), we can state that the ranges are 0 to 3 and 5 respectively we have

$0 \leq x \leq 2 \ and \ 0 \leq y \leq 5$

Using strokes theorem to evaluate

$F = z2i + 5xyj + 2y2k$

$curl \ F = 4yi+2zj+5yk = (4y,2z,5y)$

$curl \ F * nds= \frac{1}{5}(-2z + 25y)\ dy \ dx$

In conclusion, The work done is

$W=\int _CF *dr$

$\int _C F * dr = \int \int curl \ F * n ds$

$\int _C F *dr = \frac{123}{2}\int_0^3 \ dx$

$\int _C F * dr = \frac{123*3}{2}$

W= 184.5units