What is the result when 2x^3 -9x^2 +11x-6 is divided by x-3 @michele_laino

3 I   2    -9    11   -6
   I   
   I   2     6    -9     6
   
       2    -3     2     0

2x² - 3x + 2 is the answer

2x^2 - 3x + 2  
I hope that the formatting isn't destroyed since this is an exercise in long division of polynomials. We have:
 (2x^3 - 9x^2 + 11x - 6)/(x-3) 
 Look at the highest term in the numerator which is 2x^3. Now divide by the highest term in the divisor which is x. Perform that division, so 2x^3/x = 2x^2. And finally, multiply the divisor by the quotient term you just made and then subtract that from the numerator. So we have
 Partial quotient = 2x^3/x = 2x^2
 Multiply the divisor = (x-3)*2x^2 = 2x^3 - 6x^2
 Subtract from numerator = (2x^3 - 9x^2 + 11x - 6) - (2x^3 - 6x^2)
 = (- 9x^2 + 11x - 6) + 6x^2
 = (- 3x^2 + 11x - 6) 
 So we now have a quotient of 2x^2 and a simpler problem of
 (- 3x^2 + 11x - 6)/(x-3). Let's continue. 
 Partial quotient = -3x^2/x = -3x
 Multiply the divisor = -3x(x-3) = -3x^2 + 9x
 Subtract from numerator = (- 3x^2 + 11x - 6) - (-3x^2 + 9x)
 = (11x - 6) - 9x
 = 2x - 6 
 And our quotient so far is now 2x^2 - 3x, and the simpler problem is: (2x -
6)/(x - 3). So let's continue:
 Partial quotient = 2x/x = 2
 Multiply the divisor = 2(x-3) = 2x - 6
 Subtract from numerator = (2x - 6) - (2x - 6) = 0 
 And we're done with the quotient of 2x^2 - 3x + 2

Okay so for every triangle has three angles that always add up to 180 degrees. and to tell the difference between whether they are secure, right or obtuse is that acute: has one angle that is less than 90 degrees. right triangle means there is at least one angle that is exactly 90 degrees. and an obtuse angle has an angle that is bigger than 180 degrees. i’ll do #13 as an example: 13. 40 and 50 so if every triangle has 180 degrees 40+50= 90 degrees then180-90= 90 degrees so the third angle is 90 degrees and because this triangle has a 90 degree angle it means it is a right triangle