Square OABC is drawn on a centimetre grid. O is (0,0) A is (3,0) B is (3,3) C is (0,3) Write down how many invariant points there are on the perimeter of the square when OABC is translated by the vector [1 3]

No invariant point

Step-by-step explanation:

Hello!

When we translate a form, in this case a polygon We must observe the direction of the vector. Since our vector is:

1) Let's apply  that translation to this polygon, a square. Check it below:

2) The invariant points are the points that didn't change after the transformation, simply put the points that haven't changed.

Examining the graph, we can see that no, there is not an  invariant point, after the translation. There is no common point that belongs to OABC and O'A'B'C' simultaneously. All points moved.

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."

step-by-step explanation: