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: 9/8
81:72
Simplified: 9/8
B
Step-by-step explanation:
Area of square base:
13 x 13 = 169
Area of triangular sides:
B x H/2
13 x 16/2 = 104
104 x 4 = because there are 4 triangular sides of the figure = 416
416 + 169 = 585in^2
590in^2
Omg i need help on tht too
