What is the remainder when 2x^3−9x^2+11 is divided by x−6?
1 answer:
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Answer:
Approximately 17.88x or
Step-by-step explanation:
Use pythagorean formula. In a rhombus the diagonals bisect each other and they are perpendicular, so you could have a right triangle with legs of 2x and 4x, the hypoteneuse would then be which is approximately 4.47x. In a rhombus all 4 sides are the same, so multiply that by 4 and you get the perimeter. 4(4.47x) = 17.88x or if you simplify the radical instead it's
x² + 50
= x² - (-50)
= Side note:
= (x + 5i)(x - 5i)