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).
Its not clearly given that whether EBF = 2x + 9 or 2x - 9.
I have written the solution for both.
If EBF = 2x + 9,
then ABF = 6x + 26 and ABE = 11x - 31.
Now, ABE = ABF + EBF
11x - 31 = (6x + 26) + (2x + 9)
= (6x + 2x) + (26 + 9)
= 8x + 35
11x - 8x = 35 + 31
3x = 66
x = 22
Therefore, ABF = 6x + 26 = 6(22) + 26 = 132 + 26 = 158°
If EBF = 2x - 9,
then ABF = 6x + 26 and ABE = 11x - 31.
Now, ABE = ABF + EBF
11x - 31 = (6x + 26) + (2x - 9)
= (6x + 2x) + (26 - 9)
= 8x + 17
11x - 8x = 17 + 31
3x = 48
x = 16
Therefore, ABF = 6x + 26 = 6(16) + 26 = 96 + 26 = 122°
As we look, line TU = 16. But, there are two lines right below it, and there are two lines on the line UW. Now, all that this means is UW is the same length as TU.
Since we know what TU and UW is and we need to add them together to find the total length of the line TW.
16 + 16 = 32.
Therefore, the length of line TW is 32 units long.
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I got a. I multiplied 0.125 by 45 and got 5.625. I then divided 5.625 by five an got 1.125. I'm not really sure about my answer, so I did it another way. I divided .125 by five and got .025 so I multiplied that by 45 and got 1.125. I multiplied 1.125 by 1000 and got A.
Answer:
it's false
the true is -1/2
Step-by-step explanation:
-5/2-2+4=-1/2
