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statuscvo [17]
3 years ago
7

Find rational numbers between

Mathematics
1 answer:
iris [78.8K]3 years ago
5 0

Answer:

i know there's a lot of explanation. but it helps u for sure :)

Step-by-step explanation:

1)

-7 \ and \ \frac{1}{3} = \frac{-7}{1} \ and \ \frac{1}{3}\\\\LCM \ of \ 1 \ and\ 3 = 3\\\\\frac{-21}{3} \ and \ \frac{1}{3}\\\\To \ find \ rational \ numbers \ between \ \frac{-21}{3} \ and \ \frac{1}{3} \ write \ any \ number \ between \ -21 \ and \ 1 \ with \ denominator \ 3. \\\\That \ is, \  \frac{-20}{3}, \frac{-19}{3}, \frac{-18}{3}.....

2)

\frac{5}{9} \ and \ \frac{2}{3}\\\\Similarly \ take \ LCM \ of \ 9 \ and \ 3 = 9\\\\Since \ it \ is \ still \ complicated \ to \ find \ rational \ number \ between \ \frac{5}{9} \ and \ \frac{6}{9},

because \ there \ exists \ no\ natural \ number \ between \ 5 \ and \ 6.

We \ will \ multiply  \ numerator \ and \ denominator\ by\ 10. \\\\Therefore\  \frac{5}{9} \ and \ \frac{6}{9} \ becomes \ \frac{50}{90} \ and \  \frac{60}{90}.

Keeping \ denominator \ 90 \  write \ numbers \ from \  50 \ to  \ 60 \ in \ the\ numerator.\\\\That \ is , \frac{51}{90}, \frac{52}{90}, \frac{53}{90}, \frac{54}{90}, .\  .\  .

3)

LCM \ of \ 5 \ and \ 7 = 35\\\\\frac{-2}{5} \ and \ \frac{-3}{7}\  becomes \ \frac{-14}{35} \ and \ \frac{-15}{35}\\\\Now \ multiply \ denominator \ and \ numerator \ by \ 10\\\\\frac{-140}{350} \ and \ \frac{-150}{350}.\\\\Rational \ numbers  \ are \frac{-141}{350}, \frac{-142}{350}, \frac{-143}{350}, . \ . \ . \

Tip :

1. Make the denominator same.

2. Multiply numerator and denominator by 10 , 100 or 1000

3. Just write the natural numbers between the 2 numerators keeping denominator same.

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3 years ago
if a 42.9ft tall flagpole casts a 253.1ft long shadow then how long is the shadow that a 6.2ft tall woman casts?
vovangra [49]
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The answer to your question is: 
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Although you didn't ask for the solutions to the equation, I'm already here
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andreev551 [17]
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simplify and solve to find the width then you can use that to find the other values as well.
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3 years ago
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