The sum of four times a number and eight is between zero and twelve
2 answers:
Answer:
Step-by-step explanation:
The sum of four times a number and eight is between zero and twelve
For example : x is between a and b can be written as a<x<b
Now we frame inequality for the given statement
Sum of four times a number and eight
LEt the unknown number be x
four times a number is 4x
Sum of 4x and 8 is
4x+8 lies between 0 and 12. So the inequality becomes
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<span>-6x^4y^5 - 15x^3y^2 + 9x^2y^3
</span>-6x^4y^5 = -2, 3, x, x, x, x, y, y, y, y, y
15x^3y^2 = -5, 3 x, x, x, y, y
9x^2y^3 = 3, 3 x, x, y, y, y
Each group has a 3 in common and each group has 2 x in common and each group has 2 y in common so the GCF = -3x^2y^2
Divide that out and we get
</span>-6x^4y^5 = -2, 3, x, x, x, x, y, y, y, y, y
15x^3y^2 = -5, 3 x, x, x, y, y
9x^2y^3 = 3, 3 x, x, y, y, y
Each group has a 3 in common and each group has 2 x in common and each group has 2 y in common so the GCF = -3x^2y^2
Divide that out and we get