All categories

4 years ago

How can you decompose the composite figure to determine its area?

Mathematics

2 answers:

lianna [129]4 years ago

8 0

Answer is the last one (D)
semicircle, trapezoid, and 2 rectangles.

NARA [144]4 years ago

8 0

Answer:

D. As a semicircle, a trapezoid and two rectangles.

Step-by-step explanation:

Please find the attachment.

We have been given a composite figure. We are asked to decompose our given figure to determine its area.

Upon looking at attachment for our given composite figure, we can see that it consists one semicircle on top, one trapezoid in middle and two rectangles on bottom. So we can decompose our composite figure as a semicircle, a trapezoid and two rectangles.

Therefore, option D is the correct choice.

You might be interested in

Mr.santos plants beans and squash in the area shown in the model below which expression is equivalent to the sum of the two area

Mazyrski [523]

Answer:

I would go with A. 3(4 + 7) acres

Step-by-step explanation:

Easy, the expression is equivalent to the sum of two areas... Beans and the Squash... you just add them all together and you get 33 acres. Add the parenthesis first and then multiply by 3 you get 33

7 0

3 years ago

Read 2 more answers

What is the answer 7,8,9,10

stepan [7]

7) 116.6666666≈116.67
8) 138.6666666≈138.67
9) 42.1428571
10) 64.6666666≈64.67

3 0

4 years ago

Read 2 more answers

How do i go about identifying the slope & y-intercept in y = -3/4 x -7?

choli [55]

Y=mx+b the m is your slope. Here the slope would be -3/4. The b is your y-intercept which is -7.

5 0

3 years ago

Solve the inequality: −34 < −2(4x − 1)

Ne4ueva [31]

$- 34 < - 2(4x - 1) \\ - 34 < - 8x + 2 \\ - 34 - 2 < - 8x \\ - 36 < - 8x \\ 36 > 8x \\ \frac{36}{8} > x \\ x < \frac{9}{2}$

7 0

3 years ago

Please answer ASAP. The question is down below.

Butoxors [25]

Answer: 1) c 2) d

<u>Step-by-step explanation:</u>

Use Pythagorean Theorem to find the hypotenuse (r): x² + y² = r²

Given: x = 2, y = 6

2² + 6² = r²

4 + 36 = r²

40 = r²

$\sqrt{40}=r$

$\sin \theta=\dfrac{y}{r}\quad =\large\boxed{\dfrac{6}{\sqrt{40}}}\quad \rightarrow \dfrac{6}{2\sqrt{10}}\quad \rightarrow \dfrac{3}{\sqrt{10}}$

*******************************************************************************************

$\sin \theta = \dfrac{opposite(y)}{hypotenuse(r)}\quad =\dfrac{12}{20}\quad \rightarrow \dfrac{3}{5}\quad \rightarrow \large\boxed{0.6}\\\\\\\cos \theta = \dfrac{adjacent(x)}{hypotenuse(r)}\quad =\dfrac{16}{20}\quad \rightarrow \dfrac{4}{5}\quad \rightarrow \large\boxed{0.8}$

5 0

3 years ago