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baherus [9]
2 years ago
14

Question 4 (Essay Worth 10 points) (06.04 HC) 0 A figure is located at (2, 0), (2, -2), (6, -2), and (6, 0) on a coordinate plan

e. What kind of 3-D shape would be created if the figure was rotated around the x-axis? Provide an explanation and proof of your answer to receive full credit. Include the dimensions of the 3-D shape in your explanation.​
Mathematics
1 answer:
Vika [28.1K]2 years ago
7 0

The 3-D shape that would be formed is a cylinder

<h3>What kind of 3-D shape would be created if the figure was rotated around the x-axis?</h3>

The coordinates are given as:

(2, 0), (2, -2), (6, -2), and (6, 0)

From the above coordinates, we can see that the shape has horizontal and vertical lines.

This is so because the x or y axes of adjacent points are the same

So, the shape is a rectangle or a square

When a rectangle or a square is rotated around the x-axis, the 3-D shape that would be formed is a cylinder

Hence, the 3-D shape that would be formed is a cylinder

Read more about 3-D shapes at:

brainly.com/question/24383853

#SPJ1

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What is the value of y?
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Answer: y = 16.5

Step-by-step explanation:

We would apply Pythagoras theorem which is expressed as

Hypotenuse² = opposite side² + adjacent side²

Considering triangle BCD,

BD² = 14² - 8²

BD² ° 196 - 64 = 132

BD = √132

Considering triangle ABD,

AB² = BD² + y²

AB² = (√132)² + y²

AB² = 132 + y²- - - - - - - - - - 1

Considering triangle ABC,

(8 + y)² = AB² + BC²

(8 + y)² = AB² + 14²

(8 + y)² = AB² + 196

AB² = (8 + y)² - 196- - - - - - - -2

Substituting equation 1 into equation 2, it becomes

132 + y² = (8 + y)² - 196

y² - (8 + y)² = - 196 - 132

y² - (64 + 16y + y²) = - 328

y² - 64 - 16y - y2 = - 328

- 16y = - 328 + 64

- 16y = - 264

y = - 264/- 16

y = 16.5

6 0
3 years ago
In the diagram, ABCD is a part of a right angled triangle ODC. If /AB/= 6 cm,
Andreas93 [3]

Answer:

1) 13.3 cm

2) 48.3 cm

Step-by-step explanation:

1) Right angled triangle ODC and right angled triangle OAB are similar because AB//DC. The two triangles have the same proportion and are equiangular (having equal angles) but have different lengths.

Let OB = x, OC = OB + BC = x + 8

Therefore:

\frac{OC}{OB}=\frac{DC}{AB}  \\\frac{x+8}{x}=\frac{15}{6}\\15x=6(x+8)\\15x=6x+48\\15x-6x=48\\9x=48\\x=5.3\ cm

The height of triangle ODC = OC =  x + 8 = 5.3 + 8 = 13.3 cm

2) Using Pythagoras theorem:

OD² = OC² + DC²

OD² = 13.3² + 15²

OD² = 401.89

OD = √401.89 = 20 cm

2) perimeter of triangle ODC = OD + OC + DC = 20 + 13.3 + 15 =48.3 cm

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Answer:

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Answer:

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