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Sagot :
Answer:
7.3333333
Step-by-step explanation:
1 2/3 - (- 5 2/3) = 22/3 = 7 1/3= 7.3333333
detailed:
Conversion a mixed number 1 2/3
to a improper fraction: 1 2/3 = 1 2/3 = 1 · 3 + 2/3 = 3 + 2/3 = 5/3
To find a new numerator:
a) Multiply the whole number 1 by the denominator 3. Whole number 1 equally 1 * 3/3 = 3/3
b) Add the answer from previous step 3 to the numerator 2. New numerator is 3 + 2 = 5
c) Write a previous answer (new numerator 5) over the denominator 3.
One and two thirds is five thirds
Conversion a mixed number -5 2/3
to a improper fraction: -5 2/3 = -5 2/3 = -5 · 3 + (-2)/3 = -15 + (-2)/3 = -17/3
To find a new numerator:
a) Multiply the whole number -5 by the denominator 3. Whole number -5 equally -5 * 3/3 = -15/3
b) Add the answer from previous step -15 to the numerator 2. New numerator is -15 + 2 = -13
c) Write a previous answer (new numerator -13) over the denominator 3.
Minus five and two thirds is minus thirteen thirds
Subtract: 5/3 - (-17/3) = 5 - (-17)/3 = 22/3
For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(3, 3) = 3. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 3 × 3 = 9. In the following intermediate step, it cannot further simplify the fraction result by canceling.
In other words - five thirds minus minus seventeen thirds = twenty-two thirds.
Given Equation:-
[tex] \boxed{ \frak{1 \tt \frac{2}{3} - 5 \frac{2}{3}}}[/tex]
[tex] \\ [/tex]
Steps:-
[tex] \dashrightarrow\sf1 \dfrac{2}{3} - 5 \dfrac{2}{3}[/tex]
[tex] \\ [/tex]
[tex] \dashrightarrow\sf\dfrac{5}{3} - 5 \dfrac{2}{3}[/tex]
[tex] \\ [/tex]
[tex] \dashrightarrow\sf\dfrac{5}{3} - \dfrac{17}{3}[/tex]
[tex] \\ [/tex]
[tex] \dashrightarrow\sf\dfrac{5 - 17}{3} [/tex]
[tex] \\ [/tex]
[tex] \dashrightarrow \underbrace{\bold{\dfrac{ - 12}{3} } }_{ \frak{ \purple{required \: \pink{answer}}}}[/tex]
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