Using the transformation T : (x, y) → (x + 2, y + 1), find the distance named Find C'A'
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Have you learned matrices yet? I'm going to use that to solve these, please refer to the photos.
I solved the system by using a matrix calculation. The work always needs to be shown and you do that by setting up the matrix like in the first photo and also writing out all your equations. In a TI calculator, to do this you press 2ND-X^-1,GO TO EDIT - [A] - 3×4- now enter the coefficient of the system of equations variables. NOW 2ND MODE- 2ND-X^-1 -GO TO MATH- ALPHA APPS
Now, if you don't know this you may be confuse so after rref([A]) was enter a matrix came up with [1 0 0 -21] in the top row this means x equal -21. So, the next row is y and it came out as [0 1 0 46] so y equals 46 and I'm going to let you figure out what z is by looking at the matrix.
SO... X=-21 Y=46 and Z=-10
Let's Test it
I solved the system by using a matrix calculation. The work always needs to be shown and you do that by setting up the matrix like in the first photo and also writing out all your equations. In a TI calculator, to do this you press 2ND-X^-1,GO TO EDIT - [A] - 3×4- now enter the coefficient of the system of equations variables. NOW 2ND MODE- 2ND-X^-1 -GO TO MATH- ALPHA APPS
Now, if you don't know this you may be confuse so after rref([A]) was enter a matrix came up with [1 0 0 -21] in the top row this means x equal -21. So, the next row is y and it came out as [0 1 0 46] so y equals 46 and I'm going to let you figure out what z is by looking at the matrix.
SO... X=-21 Y=46 and Z=-10
Let's Test it
There is 2 figure here, a triangle and a parallelogram.
Area of parallelogram=base * height
A=9 * 5
A=45m²
Area of triangle=
(base * height)
A= (9*8)
A=72
A=36m²
36+45=81m²
The area of this figure is 81m²
Area of parallelogram=base * height
A=9 * 5
A=45m²
Area of triangle=
A=
A=
A=36m²
36+45=81m²
The area of this figure is 81m²