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3 years ago

How do you find the area of a irregular shape

2 answers:

6 0

You need to split up the irregular shape into shapes that you are familiar with them find the area of each piece and add it together. i’m not sure if that made sense.

Anna35 [415]3 years ago

5 0

To find the area of irregular shapes, the first thing to do is to divide the irregular shapeinto regular shapes that you can recognize such as triangles, rectangles, circles, squares and so forth. Then, find the area of these individual shapes and add them up!

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Sketch the region that is common to the graphs of x ≥ 2,

Oliga [24]

a. Find the graph of their common region in the attachment

b. The area of the common region of the graphs is 8 units²

<h3>a. How to sketch the region common to the graphs?</h3>

Since we have x ≥ 2, y ≥ 0, and x + y ≤ 6, we plot each graph separately and find their region of intersection.

The graph of x ≥ 2 is the region right of the line x = 2.

The graph of y ≥ 0 is the region above the line y = 0 or x-axis.

To plot the graph of x + y ≤ 6, we first plot the graph of x + y = 6 ⇒ y = -x + 6. Then the graph of x + y ≤ 6 is the graph of y ≤ - x + 6.

So, the graph of x + y ≤ 6 is the region below the line y = - x + 6

From the graph, the regions intersect at (2, 0), (2, 4) and (6, 0)

Find the graph of their common region in the attachment

<h3>b. The area of the common region</h3>

From the graph, we see that the common region is a right angled triangle with

height = 4 units and
base = 4 units

So, its area = 1/2 × height × base

= 1/2 × 4 units × 4 units

= 1/2 × 16 units²

= 8 units²

So, the area of the common region of the graphs is 8 units²

Learn more about region common to graphs here:

brainly.com/question/27932405

#SPJ1

6 0

2 years ago

How to find surface area of a triangular prism

Marrrta [24]

Here is the formula: b•h/2

3 0

4 years ago

PLEASE HELP!

Wittaler [7]

Answer:

See below

Step-by-step explanation:

Use the formulae directly

For a cone, with base radius = r and height = h, here are the related formula

$\textrm{Slant height l} = \sqrt{r^2 + h^2}$ (1)

$\textrm{Lateral surface area} = \pi r l$

$\textrm {Base area} = \pi r^2$

$\textrm { Total Surface Area SA} = \textrm{Base Area + Lateral Area} = \pi r^2 + \pi rl = \pi r(r + l)$ (2)

$\textrm{Volume V = } (1/3) \pi r^2h$ (3)

Therefore directly plugging in the numbers in the above equations:

Note:

l = slant height in cm
SA = total surface area in sqcm
V = Volume in cubic cm

Figure(a)

r = 4, h = 8

$\textrm{l} = \sqrt{4^2 + 8^2} =\sqrt{80} = 8.944 \\\textrm{SA} = 4\pi(4 + 8.944) = 4\pi(12.944) = 162.66\\\textrm{V} = (1/3)\pi(4^2)(8) = 134.04$

Figure(b)

r = 7, h =15

$\textrm{l} = \sqrt{7^2 + 15^2} =\sqrt{274} = 16.55$

$\textrm{SA} = 7\pi(7 + 16.55) = 517.89$

Figure (c)

r = 5, l = 8

$h = \sqrt{l^2 - r^2} = \sqrt{8^2 - 5^2} = \sqrt{39} = 6.245\\SA = \textrm{SA} = 5\pi(5 + 8) = 204.2\\V = (1/3)\pi(5^2)(6.245) = 163.5$

6 0

2 years ago

Consider the quadratic functions represented below. Function #1 Function #2 x y -1 14 -0 6 1 2 2 2 3 6 4 14 Which function has a

Harrizon [31]

Answer:

Function <u>#2</u> has a greater minimum.

#3 < #1 < #2

Step-by-step explanation:

In the picture attached, the question is shown.

The minimum of Function #1 is located at (3, -1). This is seen in the picture.

The minimum of Function #2 is located at (1.5, 1). We can see in the table that the function is symmetric respect 1.5 (half-point between 1 and 2).

The function y = x² + 3x - 4 has its minimum at its vertex:

x-coordinate of vertex: x = -b/(2a) = -3/(2*1) = -1.5

y-coordinate of vertex: y = (-1.5)² + 3(-1.5) - 4 = -6.25

So, the minimum of Function #3 is located at (-1.5, -6.25)

4 0

4 years ago

Exercise 30 please , show work

ArbitrLikvidat [17]

$-8(2a-3b)-5(6b-4a)=\\=-8 \times 2a -8 \times (-3b)-5 \times 6b-5 \times (-4a)= \\
=-16a+24b-30b+20a= \\
=20a-16a+24b-30b= \\
=(20-16)a+(24-30)b= \\
=4a-6b$

3 0

4 years ago

Read 2 more answers