Answer:
The image of the point (1, -2) under a dilation of 3 is (3, -6).
Step-by-step explanation:
Correct statement is:
<em>What are the coordinates of the image of the point (1, -2) under a dilation of 3 with the origin.</em>
From Linear Algebra we get that dilation of a point with respect to another point is represented by:
(Eq. 1)
Where:
- Reference point with respect to origin, dimensionless.
- Original point with respect to origin, dimensionless.
- Dilation factor, dimensionless.
If we know that 
, 
and 
, then the coordinates of the image of the original point is:
![\vec P' = (0,0) +3\cdot [(1,-2)-(0,0)]](https://tex.z-dn.net/?f=%5Cvec%20P%27%20%3D%20%280%2C0%29%20%2B3%5Ccdot%20%5B%281%2C-2%29-%280%2C0%29%5D)
The image of the point (1, -2) under a dilation of 3 is (3, -6).
Answer:
f [ x ] = [ x - 3] [ x + 8 ] = x^2 + 5x - 24
Step-by-step explanation:
f (x)=[ x - 3] [ x + 8 ]
x [ x + 8 ] -3[ x + 8 ]
x^2 + 8x - 3x - 24
x^2 + 5x - 24
Answer:
angle BED and angle DEB are the answers.
Answer:
x=58, m<DCB=56°
Step-by-step explanation:
Ok the sum of those two angles should equal 180 degrees so,
2x+8+x-2=180
solve the equation
3x+6=180
3x=174
x= 58
M<DCB=x-2--> 58-2=56
Answer:
U = -5
Step-by-step explanation:
-8u = -7u + 5
-u = 5
u = -5
