Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
Secondary 1 | Maths
2 Answers Below
Anyone can contribute an answer, even non-tutors.
Pls help
Date Posted: 1 year ago
Views: 456
See 2 Answers
8a) Yes, but only under the circumstances where m & n are multiples of 2 (even numbers), hence it is possible to divide both m & n by 2 completely and form a whole number of 2cm x 2cm squares with no remainder.
8b) To find the number of ways to split this rectangle into squares, we need to find the number of common factors between 280 and 145:
280=1x2x2x2x5x7
145=1x5x29
As such these 2 numbers only have 2 commin factors, 1 and 5, and it is only possible to split this rectangle into squares in 2 ways.
8b) To find the number of ways to split this rectangle into squares, we need to find the number of common factors between 280 and 145:
280=1x2x2x2x5x7
145=1x5x29
As such these 2 numbers only have 2 commin factors, 1 and 5, and it is only possible to split this rectangle into squares in 2 ways.
Thank you soo much
9a) 1176 can be divided by 2, as it is an even number,and can be divided by 4, as 76 is a multiple of 4
1176 ÷ 4 = 294
294 can be further divided by 2
294 ÷ 2 = 147
147 can be further divided by 3, as the sum of its digits are a multiple of 3:
147÷3=49
49 also equal to 49=7x7
as such 1176=4x2x3x7x7=2³x3¹x7²
hence, a=3, b=1 and c=2
9bi)
1176=2x2x2x3x7x7
if the HCF of 1176 and m is 2x2x3x7
Since m is a multiple of 2x2x3x7, and that is already a factor of 1176, then Smallest Value of M=2x2x3x7
9bii) For another possible value of M, it will have a to be a multiple if 2x2x3x7 but have a factor that is not one of the prime factors of 1176 (2, 3 or 7). For example 5, or 11, or 13, as such, m can also be
a) m=5x2x2x3x7
b) m=11x2x2x3x7
c) m=13x2x2x3x7
1176 ÷ 4 = 294
294 can be further divided by 2
294 ÷ 2 = 147
147 can be further divided by 3, as the sum of its digits are a multiple of 3:
147÷3=49
49 also equal to 49=7x7
as such 1176=4x2x3x7x7=2³x3¹x7²
hence, a=3, b=1 and c=2
9bi)
1176=2x2x2x3x7x7
if the HCF of 1176 and m is 2x2x3x7
Since m is a multiple of 2x2x3x7, and that is already a factor of 1176, then Smallest Value of M=2x2x3x7
9bii) For another possible value of M, it will have a to be a multiple if 2x2x3x7 but have a factor that is not one of the prime factors of 1176 (2, 3 or 7). For example 5, or 11, or 13, as such, m can also be
a) m=5x2x2x3x7
b) m=11x2x2x3x7
c) m=13x2x2x3x7
360 Tutoring Program