The HCF of two numbers is 20. The LCM of both numbers is a multiple of 14. One number is greater than the other, work out the sm
allest possible numbers
1 answer:
Answer:
20
Step-by-step explanation:
Given that the HCF of two numbers is 20 and the LCM of both numbers is a multiple of 14.
So, the LCM of both numbers = 14n where n can be any natural number.
As HCF of two numbers is always smaller than or equal to the smallest number among both the numbers.
Here, the HCF of two numbers is 20, so the smallest number is greater than of equal to 20.
Hence, the smallest possible number is 20.
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c
Step-by-step explanation:
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Answer:
16/63
Step-by-step explanation:
8/9 * 2/7
Multiply the numerators
8*2 = 16
Multiply the denominators
9*7 = 63
Put the numerator over the denominator
16/63
The sequence is a linear function with a slope of 1 and a y intercept of -2. The only choice that meets these conditions is a=n-2.
#1
- 15x²+4x-4
- 15x²+10x-6x-4
- 5x(3x+2)-2(3x+2)
- (5x-2)(3x+2)
#2
- 16a²-50a+36
- 2(8a²-25a+18)
- 2(8a-9)(a-2)
#3
- 63n²+126n+48
- 3(21n²+42n+16)
Not more factorable
Answer:
27a³
Step-by-step explanation:
9⁴xa⁷b⁴ / 3⁵xa⁴b⁴
(3²)⁴xa⁷b⁴ / 3⁵xa⁴b⁴
3⁸xa⁷b⁴ / 3⁵xa⁴b⁴
3³a³
27a³