Jake's Game of Questionable Quality [6/5+]

For completed/abandoned Mish Mash Games.
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Post Post #100 (ISO) » Wed Jan 26, 2022 12:06 pm

Post by Felissan »

Check inventory, while lovingly incubating the egg
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Post Post #101 (ISO) » Wed Jan 26, 2022 12:10 pm

Post by Save The Dragons »

feed the inch dragon bath salts
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Post Post #102 (ISO) » Wed Jan 26, 2022 12:22 pm

Post by Jake The Wolfie »

In post 100, Felissan wrote:Check inventory, while lovingly incubating the egg
Felissan has earned an egg piece!

You currently have:

  • 4-Shell Egg (x1)
  • Egg piece (x1)
  • PokeCube™
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Post Post #103 (ISO) » Wed Jan 26, 2022 12:23 pm

Post by Jake The Wolfie »

In post 101, Save The Dragons wrote:feed the inch dragon bath salts
Inch Dragon is no longer Hungry and is now Inch Inch Dragon
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Post Post #104 (ISO) » Wed Jan 26, 2022 12:23 pm

Post by Save The Dragons »

play with inch inch dragon
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Post Post #105 (ISO) » Wed Jan 26, 2022 12:31 pm

Post by Jake The Wolfie »

In post 104, Save The Dragons wrote:play with inch inch dragon
Inch Inch Dragon is sleepy.
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Post Post #106 (ISO) » Wed Jan 26, 2022 12:51 pm

Post by Ircher »

HEAL: Save the Dragons

Inspect inventory.

Steal paper and pencil.
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Post Post #107 (ISO) » Wed Jan 26, 2022 2:46 pm

Post by StrangerCoug »

Help Save the Dragons find Inch Inch Dragon somewhere to sleep
STRANGERCOUG: Stranger Than You!

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Post Post #108 (ISO) » Wed Jan 26, 2022 10:34 pm

Post by Felissan »

Poke the PokeCube™
"Dammit Felissan, making someone lose the game is NOT NICE"
- DeathRowKitty 2016
"Also, the me in your signature just made the me in this thread lose the game and I'm not sure how to feel about this."
- DeathRowKitty 2018
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Post Post #109 (ISO) » Thu Jan 27, 2022 6:59 am

Post by Jake The Wolfie »

In post 106, Ircher wrote:HEAL: Save the Dragons

Inspect inventory.

Steal paper and pencil.
Save The Dragons is no longer poisoned
When Stealing the Pencil and Paper, you are caught.
You currently have:
  • Egg Piece (x1)
  • Thirty Sevens (Casino)
  • Pencil and Paper (Stolen)
  • A ruler and compass and calculator all-in-one.
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Post Post #110 (ISO) » Thu Jan 27, 2022 7:00 am

Post by Jake The Wolfie »

In post 107, StrangerCoug wrote:
Help Save the Dragons find Inch Inch Dragon somewhere to sleep
Save The Dragon's Inch Inch Dragon is sleepy
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Post Post #111 (ISO) » Thu Jan 27, 2022 7:01 am

Post by Jake The Wolfie »

In post 108, Felissan wrote:Poke the PokeCube™
The cube seems indifferent to being poked
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Post Post #112 (ISO) » Thu Jan 27, 2022 7:37 am

Post by StrangerCoug »

Turn laptop back on
Find bank to hack into
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Post Post #113 (ISO) » Thu Jan 27, 2022 8:52 am

Post by Ircher »

Construct the square root of 14 using the compass and ruler on a piece of paper by following the instructions given below and keep the construction lines:

Spoiler: Construction Instructions
I highly encourage you to actually try this using a tool like geogebra.

Part I: Constructing the square root of 2.
1: Start with a line segment AB and extend it to be a line.
2: Draw a circle with center A and radius AB. Label the point D as the other point on the circle that intersects the line.
3: Draw a circle with center D and radius BD.
4: Draw a circle with center B and radius BD.
5: Label the two intersection points from steps 4 and 5 as E and F.
6: Connect the line segment EF. Label one of the intersection point on EF with the circle from step 3 as C. Notice that steps 3-6 constructed a line EF that is perpendicular to the line AB.
7: Connect points B and C to form the segment BC. By the Pythagorean Theorem, the hypotenuse of triangle ABC has length equal to the square root of 2 (given the segment AB has length 1).

Part II: Constructing the square root of 3 from the square root of 2.
8: Extend the line segment BC to be a line.
9: Using a process similar to steps 2-6 above, create the line perpendicular to the line BC that intersects BC at the point B. Instructions 9a through 9d give a reminder of the process.
9a: Draw a circle with center B and radius BC. Label the point that intersects the line BC as G.
9b: Draw a circle with center C and radius CG.
9c: Draw a circle with center G and radius CG.
9d: Draw the line segment that intersects the two circles. This line is perpendicular to BC and intersects the point B.
10: Using a compass, measure the length AB. Then draw a circle with radius AB and center B.
11: Mark the point where the circle from step 10 intersects the line segment from step 9 as B2. The line segment should have length AB.
12: Connect points B2 and C to form triangle CBB2. By the Pythagorean Theorem, the line CB2 has length proportional to the square root of 3.

Part III: Constructing the square root of n in general:

13: Construct the square root of n - 1. Note that the point C should be shared by the hypotenuses of each triangle formed up to this point. The vertex of the right angle in the last triangle that constructs the square root of n - 1 is Bn - 3 and the remaining vertex is Bn - 2. (Note that point B should be considered point B1.)
14: Construct the line perpendicular to CBn - 2 that intersects the point Bn - 2. Refer to steps 2-6 or 9 for a reminder of the process.
15: Use the compass to measure the length AB. Draw a circle with radius AB and center Bn - 2.
16: Label the point where the circle from step 15 intersects the line from step 14 as the point Bn - 1.
17: Connect points C and Bn - 1. By the Pythagorean Theorem, this line segment has length proportional to the square root of n.


Sell it as a piece of art.
Links: User Page | GTKAS
Do you have questions, ideas, or feedback for the Scummies? Please pm me!
Hosting: The Grand Neighborhood [Ongoing]
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Post Post #114 (ISO) » Thu Jan 27, 2022 9:03 am

Post by Jake The Wolfie »

In post 113, Ircher wrote:Construct the square root of 14 using the compass and ruler on a piece of paper by following the instructions given below and keep the construction lines:

Spoiler: Construction Instructions
I highly encourage you to actually try this using a tool like geogebra.

Part I: Constructing the square root of 2.
1: Start with a line segment AB and extend it to be a line.
2: Draw a circle with center A and radius AB. Label the point D as the other point on the circle that intersects the line.
3: Draw a circle with center D and radius BD.
4: Draw a circle with center B and radius BD.
5: Label the two intersection points from steps 4 and 5 as E and F.
6: Connect the line segment EF. Label one of the intersection point on EF with the circle from step 3 as C. Notice that steps 3-6 constructed a line EF that is perpendicular to the line AB.
7: Connect points B and C to form the segment BC. By the Pythagorean Theorem, the hypotenuse of triangle ABC has length equal to the square root of 2 (given the segment AB has length 1).

Part II: Constructing the square root of 3 from the square root of 2.
8: Extend the line segment BC to be a line.
9: Using a process similar to steps 2-6 above, create the line perpendicular to the line BC that intersects BC at the point B. Instructions 9a through 9d give a reminder of the process.
9a: Draw a circle with center B and radius BC. Label the point that intersects the line BC as G.
9b: Draw a circle with center C and radius CG.
9c: Draw a circle with center G and radius CG.
9d: Draw the line segment that intersects the two circles. This line is perpendicular to BC and intersects the point B.
10: Using a compass, measure the length AB. Then draw a circle with radius AB and center B.
11: Mark the point where the circle from step 10 intersects the line segment from step 9 as B2. The line segment should have length AB.
12: Connect points B2 and C to form triangle CBB2. By the Pythagorean Theorem, the line CB2 has length proportional to the square root of 3.

Part III: Constructing the square root of n in general:

13: Construct the square root of n - 1. Note that the point C should be shared by the hypotenuses of each triangle formed up to this point. The vertex of the right angle in the last triangle that constructs the square root of n - 1 is Bn - 3 and the remaining vertex is Bn - 2. (Note that point B should be considered point B1.)
14: Construct the line perpendicular to CBn - 2 that intersects the point Bn - 2. Refer to steps 2-6 or 9 for a reminder of the process.
15: Use the compass to measure the length AB. Draw a circle with radius AB and center Bn - 2.
16: Label the point where the circle from step 15 intersects the line from step 14 as the point Bn - 1.
17: Connect points C and Bn - 1. By the Pythagorean Theorem, this line segment has length proportional to the square root of n.


Sell it as a piece of art.
Awful. Not even a hint of colour. You barely manage to make a one.
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Post Post #115 (ISO) » Thu Jan 27, 2022 9:05 am

Post by Jake The Wolfie »

In post 112, StrangerCoug wrote:
Turn laptop back on
Find bank to hack into
The laptop is dead.
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Post Post #116 (ISO) » Thu Jan 27, 2022 9:09 am

Post by Ircher »

HURT WITH A BLADE: Jake The Wolfie
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Post Post #117 (ISO) » Thu Jan 27, 2022 9:25 am

Post by Jake The Wolfie »

Jake The Wolfie is currently poisoned.
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Post Post #118 (ISO) » Thu Jan 27, 2022 9:54 am

Post by Ircher »

Walk to the next nearest town that isn't the one I'm currently in.
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Post Post #119 (ISO) » Thu Jan 27, 2022 10:07 am

Post by Felissan »

Ircher wrote:Construct the square root of 14 using the compass and ruler on a piece of paper by following the instructions given below and keep the construction lines:

Spoiler: Construction Instructions
I highly encourage you to actually try this using a tool like geogebra.

Part I: Constructing the square root of 2.
1: Start with a line segment AB and extend it to be a line.
2: Draw a circle with center A and radius AB. Label the point D as the other point on the circle that intersects the line.
3: Draw a circle with center D and radius BD.
4: Draw a circle with center B and radius BD.
5: Label the two intersection points from steps 4 and 5 as E and F.
6: Connect the line segment EF. Label one of the intersection point on EF with the circle from step 3 as C. Notice that steps 3-6 constructed a line EF that is perpendicular to the line AB.
7: Connect points B and C to form the segment BC. By the Pythagorean Theorem, the hypotenuse of triangle ABC has length equal to the square root of 2 (given the segment AB has length 1).

Part II: Constructing the square root of 3 from the square root of 2.
8: Extend the line segment BC to be a line.
9: Using a process similar to steps 2-6 above, create the line perpendicular to the line BC that intersects BC at the point B. Instructions 9a through 9d give a reminder of the process.
9a: Draw a circle with center B and radius BC. Label the point that intersects the line BC as G.
9b: Draw a circle with center C and radius CG.
9c: Draw a circle with center G and radius CG.
9d: Draw the line segment that intersects the two circles. This line is perpendicular to BC and intersects the point B.
10: Using a compass, measure the length AB. Then draw a circle with radius AB and center B.
11: Mark the point where the circle from step 10 intersects the line segment from step 9 as B2. The line segment should have length AB.
12: Connect points B2 and C to form triangle CBB2. By the Pythagorean Theorem, the line CB2 has length proportional to the square root of 3.

Part III: Constructing the square root of n in general:

13: Construct the square root of n - 1. Note that the point C should be shared by the hypotenuses of each triangle formed up to this point. The vertex of the right angle in the last triangle that constructs the square root of n - 1 is Bn - 3 and the remaining vertex is Bn - 2. (Note that point B should be considered point B1.)
14: Construct the line perpendicular to CBn - 2 that intersects the point Bn - 2. Refer to steps 2-6 or 9 for a reminder of the process.
15: Use the compass to measure the length AB. Draw a circle with radius AB and center Bn - 2.
16: Label the point where the circle from step 15 intersects the line from step 14 as the point Bn - 1.
17: Connect points C and Bn - 1. By the Pythagorean Theorem, this line segment has length proportional to the square root of n.


Sell it as a piece of art.
Oh, you thought you could get away with not showing the picture, didn't you?

Spoiler:
And you were right, because I made it instead.
Image
"Dammit Felissan, making someone lose the game is NOT NICE"
- DeathRowKitty 2016
"Also, the me in your signature just made the me in this thread lose the game and I'm not sure how to feel about this."
- DeathRowKitty 2018
"You've made me make myself lose the game so many times that I feel like it's an entirely new game I'm losing"
- DeathRowKitty 2022
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Post Post #120 (ISO) » Thu Jan 27, 2022 10:26 am

Post by Jake The Wolfie »

In post 118, Ircher wrote:Walk to the next nearest town that isn't the one I'm currently in.
Jake The Wolfie is currently poisoned.
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Post Post #121 (ISO) » Thu Jan 27, 2022 10:58 am

Post by Felissan »

Re-check my inventory!
"Dammit Felissan, making someone lose the game is NOT NICE"
- DeathRowKitty 2016
"Also, the me in your signature just made the me in this thread lose the game and I'm not sure how to feel about this."
- DeathRowKitty 2018
"You've made me make myself lose the game so many times that I feel like it's an entirely new game I'm losing"
- DeathRowKitty 2022
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Post Post #122 (ISO) » Thu Jan 27, 2022 11:12 am

Post by StrangerCoug »

Charge laptop
STRANGERCOUG: Stranger Than You!

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Post Post #123 (ISO) » Thu Jan 27, 2022 11:17 am

Post by Jake The Wolfie »

In post 121, Felissan wrote:Re-check my inventory!
Jake The Wolfie is currently poisoned.
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Post Post #124 (ISO) » Thu Jan 27, 2022 11:17 am

Post by Jake The Wolfie »

In post 122, StrangerCoug wrote:
Charge laptop
Jake The Wolfie is currently poisoned.
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