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Debora [2.8K]
3 years ago
14

More than one unknown in a word problem can be represented in terms of the same variable assigned to first unknown. true of fals

e
Mathematics
2 answers:
Sergeeva-Olga [200]3 years ago
6 0
This is clearly false as it will just mess up the question.
Hitman42 [59]3 years ago
5 0

Answer:

True

Step-by-step explanation:

When you have a word problem with more than one unknown data, you can represent these unknowns in terms of the same variable assigned to the first unknown (it depends on the problem).

For example, if the word problem tells us that  Alex has 3 more years than Jennifer and that Ben has twice the age of Jennifer plus 4, you can represent all these unknowns (the three ages) in terms of the variable assigned to Jennifer's age. Let's see how this would be:

If you name x the age of Jennifer, then Alex's age would be x + 3 (3 more years than Jennifer), and Ben's age would be 2x + 4 (twice the age of Jennifer plus 4).

You can do this in some word problems to simplify the process of finding your unknowns because you'd be working with just one variable instead of many of them.

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mrs_skeptik [129]

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Will mark brainiest
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40 POINTS!!!! ANSWER CORRECTLY PLZ
Grace [21]

Answer:

5. ○ \displaystyle \frac{25}{48}

4. ○ \displaystyle \frac{25}{46}

3. ○ \displaystyle \frac{31}{100}

2. ○ \displaystyle \frac{3}{4}

1. ○ \displaystyle \frac{27}{50}

Step-by-step explanation:

5. \displaystyle \frac{25}{23 + 25} = \frac{25}{48}

4. \displaystyle \frac{25}{46} = \frac{25}{21 + 25}

3. \displaystyle Flat-out\:given\:to\:you

2. \displaystyle \frac{21 + 23 + 31}{100} = \frac{75}{100} = \frac{3}{4}

1. \displaystyle \frac{31 + 23}{100} = \frac{54}{100} = \frac{27}{50}

I am joyous to assist you anytime.

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