The integers between the pairs in increasing order are: a. Between 0 and –7: –6, –5, –4, –3, –2, –1. b. Between –4 and 4: –3, –2, –1, 0, 1, 2, 3. c. Between –8 and –15: –14, –13, –12, –11, –10, –9. d. Between –30 and –23: –29, –28, –27, –26, –25, –24. Arranging integers systematically highlights theRead more
The integers between the pairs in increasing order are:
a. Between 0 and –7: –6, –5, –4, –3, –2, –1.
b. Between –4 and 4: –3, –2, –1, 0, 1, 2, 3.
c. Between –8 and –15: –14, –13, –12, –11, –10, –9.
d. Between –30 and –23: –29, –28, –27, –26, –25, –24.
Arranging integers systematically highlights their sequential placement.
Here are examples of three numbers whose sum equals –8: a. –2, –3, –3 (adds to –8). b. –4, –1, –3 (adds to –8). c. –5, 0, –3 (adds to –8). The possibilities are numerous, as the combinations depend on choosing numbers with negative values and balancing them to reach the total. Exploring more combinaRead more
Here are examples of three numbers whose sum equals –8:
a. –2, –3, –3 (adds to –8).
b. –4, –1, –3 (adds to –8).
c. –5, 0, –3 (adds to –8).
The possibilities are numerous, as the combinations depend on choosing numbers with negative values and balancing them to reach the total. Exploring more combinations provides insight into how integers interact in addition.
Using two dice with faces numbered –1, 2, –3, 4, –5, and 6, some sums between –10 and 12 cannot be achieved. The sums not possible are –9, –7, –6, –4, and 11. These gaps arise because no combination of the dice values adds to these totals. Understanding this requires analyzing all possible outcomesRead more
Using two dice with faces numbered –1, 2, –3, 4, –5, and 6, some sums between –10 and 12 cannot be achieved. The sums not possible are –9, –7, –6, –4, and 11. These gaps arise because no combination of the dice values adds to these totals. Understanding this requires analyzing all possible outcomes of adding or subtracting the dice numbers and observing which totals remain absent.
The solutions to these problems demonstrate integer operations: a. 8 – 13 = –5. Subtracting a larger number gives a negative result. b. (–8) – (13) = –21. Subtraction combines the negatives, extending the negative value. c. (–13) – (–8) = –5. Subtracting a negative is equivalent to adding a positiveRead more
The solutions to these problems demonstrate integer operations:
a. 8 – 13 = –5. Subtracting a larger number gives a negative result.
b. (–8) – (13) = –21. Subtraction combines the negatives, extending the negative value.
c. (–13) – (–8) = –5. Subtracting a negative is equivalent to adding a positive.
d. (–13) + (–8) = –21. Adding two negatives deepens the negative result. These examples illustrate key integer rules.
150 years ago from the present year (2024) was the year 1874. To find this, subtract 150 from 2024, yielding 1874. This calculation shows how subtraction helps determine historical events or milestones. Such exercises are useful in understanding timelines and connecting past events to the present. LRead more
150 years ago from the present year (2024) was the year 1874. To find this, subtract 150 from 2024, yielding 1874. This calculation shows how subtraction helps determine historical events or milestones. Such exercises are useful in understanding timelines and connecting past events to the present. Learning to compute years back accurately is essential in various subjects, including history and mathematics, where timelines often play a critical role.
Write all the integers between the given pairs, in increasing order: 0 and –7; –4 and 4; –8 and –15; –30 and –23.
The integers between the pairs in increasing order are: a. Between 0 and –7: –6, –5, –4, –3, –2, –1. b. Between –4 and 4: –3, –2, –1, 0, 1, 2, 3. c. Between –8 and –15: –14, –13, –12, –11, –10, –9. d. Between –30 and –23: –29, –28, –27, –26, –25, –24. Arranging integers systematically highlights theRead more
The integers between the pairs in increasing order are:
a. Between 0 and –7: –6, –5, –4, –3, –2, –1.
b. Between –4 and 4: –3, –2, –1, 0, 1, 2, 3.
c. Between –8 and –15: –14, –13, –12, –11, –10, –9.
d. Between –30 and –23: –29, –28, –27, –26, –25, –24.
Arranging integers systematically highlights their sequential placement.
For more NCERT Solutions for Class 6 Math Chapter 10 The Other Side of Zero Extra Questions and Answer:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-6/maths/
Give three numbers such that their sum is –8.
Here are examples of three numbers whose sum equals –8: a. –2, –3, –3 (adds to –8). b. –4, –1, –3 (adds to –8). c. –5, 0, –3 (adds to –8). The possibilities are numerous, as the combinations depend on choosing numbers with negative values and balancing them to reach the total. Exploring more combinaRead more
Here are examples of three numbers whose sum equals –8:
a. –2, –3, –3 (adds to –8).
b. –4, –1, –3 (adds to –8).
c. –5, 0, –3 (adds to –8).
The possibilities are numerous, as the combinations depend on choosing numbers with negative values and balancing them to reach the total. Exploring more combinations provides insight into how integers interact in addition.
For more NCERT Solutions for Class 6 Math Chapter 10 The Other Side of Zero Extra Questions and Answer:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-6/maths/
There are two dice with faces numbered: –1, 2, –3, 4, –5, 6. Which sums between –10 and 12 are not possible by adding the dice?
Using two dice with faces numbered –1, 2, –3, 4, –5, and 6, some sums between –10 and 12 cannot be achieved. The sums not possible are –9, –7, –6, –4, and 11. These gaps arise because no combination of the dice values adds to these totals. Understanding this requires analyzing all possible outcomesRead more
Using two dice with faces numbered –1, 2, –3, 4, –5, and 6, some sums between –10 and 12 cannot be achieved. The sums not possible are –9, –7, –6, –4, and 11. These gaps arise because no combination of the dice values adds to these totals. Understanding this requires analyzing all possible outcomes of adding or subtracting the dice numbers and observing which totals remain absent.
For more NCERT Solutions for Class 6 Math Chapter 10 The Other Side of Zero Extra Questions and Answer:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-6/maths/
Solve these: 8 – 13, (–8) – (13), (–13) – (–8), (–13) + (–8).
The solutions to these problems demonstrate integer operations: a. 8 – 13 = –5. Subtracting a larger number gives a negative result. b. (–8) – (13) = –21. Subtraction combines the negatives, extending the negative value. c. (–13) – (–8) = –5. Subtracting a negative is equivalent to adding a positiveRead more
The solutions to these problems demonstrate integer operations:
a. 8 – 13 = –5. Subtracting a larger number gives a negative result.
b. (–8) – (13) = –21. Subtraction combines the negatives, extending the negative value.
c. (–13) – (–8) = –5. Subtracting a negative is equivalent to adding a positive.
d. (–13) + (–8) = –21. Adding two negatives deepens the negative result. These examples illustrate key integer rules.
For more NCERT Solutions for Class 6 Math Chapter 10 The Other Side of Zero Extra Questions and Answer:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-6/maths/
From the present year, which year was it 150 years ago?
150 years ago from the present year (2024) was the year 1874. To find this, subtract 150 from 2024, yielding 1874. This calculation shows how subtraction helps determine historical events or milestones. Such exercises are useful in understanding timelines and connecting past events to the present. LRead more
150 years ago from the present year (2024) was the year 1874. To find this, subtract 150 from 2024, yielding 1874. This calculation shows how subtraction helps determine historical events or milestones. Such exercises are useful in understanding timelines and connecting past events to the present. Learning to compute years back accurately is essential in various subjects, including history and mathematics, where timelines often play a critical role.
For more NCERT Solutions for Class 6 Math Chapter 10 The Other Side of Zero Extra Questions and Answer:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-6/maths/