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A segment is bounded by two endpoints.
The two segments can have up to two common points
Assume the line segments are AB and CD where the length of AB is greater than the length of CD.
<u>The possibilities are:</u>
- <em>A point of segment CD lies on segment AB</em>
- <em>Both points of segment CD lie on segment AB.</em>
<em />
See attachment for both possibilities.
Hence, it is possible for the two segments to have two common points.
Read more about line segments at:
Answer:
2x² + 3x + 4 + (17x + 9) / (x² − 4)
Step-by-step explanation:
Start by setting up the division. Make sure to write all the coefficients, even the zero ones.
x² + 0x − 4 | 2x⁴ + 3x³ − 4x² + 5x − 7
Start by dividing the first term of the dividend (2x⁴) by the first term of the divisor (x²). That's 2x²; it'll be the first term in quotient. Multiply the divisor by 2x²:
2x⁴ + 0x³ − 8x²
Subtract that from the first three terms of the dividend:
3x³ + 4x²
Drop down the next term from the dividend, and start the process all over again.
3x³ + 4x² + 5x
When you finish, the quotient will be 2x² + 3x + 4, and the remainder will be 17x + 9.
