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Vladimir79 [104]
3 years ago
13

How would the following problem be represented using colored tiles? -7 + (-3) seven red tiles plus three blue tiles seven red ti

les plus three red tiles seven blue tiles plus three red tiles seven blue tiles plus three blue tiles
Mathematics
2 answers:
Lelechka [254]3 years ago
5 0

Answer:

Ok, we have the equation:

-7 + (-3)

Both numbers are negative, so the solution of this will be:

-7 + (-3) = -7 - 3 = -10.

Now, how we can represent it with tiles?

First we need to define, which tiles are the ones representing negative numbers?

We can use:

Red = positive

Blue = negative.

Then:

-7 is represented with seven blue tiles

-3 is represented with tree blue tiles

then:

(-7) + (-3)

will be represented with seven blue tiles, plus tree blue tiles.

When we add them, we have a total of 10 blue tiles, and as blue tiles represent negative numbers, 10 blue tiles is equal to -10.

murzikaleks [220]3 years ago
5 0

Answer:

-7-3

Step-by-step explanation:

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Given: ∠AOC, ∠BOC - complementary angles
34kurt

Answer:

  • 30
  • COB . . . . or . . . . BOC

Step-by-step explanation:

The reason given on the line of interest is "substitution," so the problem boils down to determining what was substituted for what. The previous statement says ...

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<em>Comment on the problem</em>

There is a curious mix of notations here. Usually, (as in the beginning of this problem) we refer to the measure of an angle using "m∠" in front of the angle designator, and we use a degree symbol to indicate the units of that measure. Part-way through the problem statement written here, those notations were dropped, and we're to assume they are intended. IMO, this is a poor way to demonstrate careful problem solving.

The substitution given for AOC is BOC+30, but the line into which that is substituted has AOC +COB = 90. This means the equation after substitution is ...

  BOC +30 +COB = 90

Since BOC and COB are the same angle, we can sort of fudge the "algebra" to get to  BOC=30, but if the problem were more carefully written, the angle would be referred to by consistent nomenclature:

  m∠AOC + m∠BOC = 90° . . . . . . . . . preferred angle designations

  (m∠BOC + 30°) + m∠BOC = 90° . . . . substitution for m∠AOC

  2(m∠BOC) = 60° . . . . . . algebra (subtract 30°, collect terms)

  m∠BOC = 30° . . . . . . . . algebra (divide by 2)

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