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Sidana [21]
3 years ago
6

Would shadows be 3 dimensional in 4 dimenional space

Mathematics
1 answer:
azamat3 years ago
4 0
4 Dimensional<span> shapes can</span><span> cast </span>3 dimensional shadows<span>. ... M</span><span>aybe a </span>shadow<span> is the fourth dimension, its been here all the time.</span>
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naomi's fish is 40 millimeters long. her guinea pig is 25 centimeters long. how much longer is the guinea pig than her fish
horrorfan [7]
Hi there!

The correct answer is 6.25 times.

This is how you get there:

1) First off, you are going to need to have the same unit (either cm or mm). I used millimeters for this problem. 25cm=250mm

2) This next part is very easy to get wrong. You need to figure out what you are dividing, and by what. Simply look at the problem. <span><em>"How much longer is the guinea pig than her fish?" </em>This shows what numbers you should be getting. The guinea pig is 250mm long, where the fish is only 40. This means that her guinea pig should be larger. 250÷40=6.25

If you did this problem the other way, you would've ended up with .16. If you put the sizes into perspective, you will notice that this is highly unlikely.

I hope that I have been able to simplify this problem for you!

Brady
</span>
3 0
3 years ago
Find the Domain and Range of the graph<br> Plz helppppppppppp
Semenov [28]
Domain: 1 2 3 4 5 6 range: -1 0 1 2 3 6
3 0
3 years ago
Read 2 more answers
3x + 8 = 9 + 3x - 14<br> What’s the answer
irina1246 [14]

Answer:

0 = -13

Step-by-step explanation:

8 0
3 years ago
Two chess players, A and B, are going to play 7 games. Each game has three possibleoutcomes: a win for A (which is a loss for B)
Lina20 [59]

Answer:

A) the possible outcomes for individual game is 210 games.

B) The possible outcome for this is 357 games.

C) Then the Possible Games are 267 games.

Step-by-step explanation:

A) Total number of individual games are 7 in which A ends up with 3 wins which give 4 remaining games. then there are two draw games from them. and in the end the remaining games are losses so,we useformula for combination:

                       (7C3)*(4C2)=210 \ games

B) Now Player A has 4 point that gives us 5 possibilities until we reach 7 games. Similarly B have 3 points which is same as player A gives us 5 possibilities until we reach 7 games. thus those two cases for player A and Player B can be stated as:

           (7C3)+(7C3)*(4C2)+(7C1)*(6C6)+(7C2)*(5C4)=357\  games

C) Lets say that player A ahs 4 points and player B has 3 points and all seven games have been played so.

c1) If player A has 4 wins and 3 losses then last win have to be in 7th match thus:the answer is (6C3).

c2) If player A has 3 win, 2 draws and 3 losses thus it means that final match cannot be a loss thus the answer is (6C2)*(5C3).

c3) Now lastly if player A has 1 win and 5 draws we can arrange them arbitrarily thus the answer here is (7C1).

c4) now If player A has 2 wins, 4 draws and 2 losses thus answer is (6C1)*(6C2)

Sum of all the cases is

               (6C3)+(6C2)*(5C3)+(7C1)+(6C1)*(6C2)=267\ games

4 0
3 years ago
3. Which ordered pair is a solution of the equation y = x – 3?
Ivan
The answer is (5,2)
7 0
3 years ago
Read 2 more answers
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