The distributive property of 27+60 ??

Which triangle results from a reflection across the line x = 1?

Nataly_w [17]

Answer:

Correct answer is option D.

Step-by-step explanation:

Given that $\triangle ABC$ in the image 1 attached.

If we have a look at the image attached, the coordinates are:

$A(1,1)\\B(2,5)\ and\\C(4,1)$

To find reflection of a point across any line, the distance of points from the line must be same.

Point A(1,1) lies on the line x = 1, so its reflection A' will be at the same point A'(1,1).

Point C(2,5) is at a distance 1 from x = 1 on right side, so C' will be 1 distance on the left side of x = 1 i.e. 1 will be subtracted from its x coordinate.

i.e. C'(1 - 1, 5)

C'(0, 5)

Point B(4, 1) is at a distance 3 from x = 1 on right side, so B' will be 3 distance on the left side of x = 1 i.e. 3 will be subtracted from its x coordinate.

i.e. B'(1 - 3, 1)

B'(-2, 1)

When we plot the above point A', B' and C', we get the option D as correct.

7 0

4 years ago

Read 2 more answers

Can somebody explain why 8/17 is correct. Will make brainliest.

irina1246 [14]

Answer:

8/17 is correct because cosine = adjacent/hypotenuse, and since angle A is theta, or the reference angle, the adjacent of A would be 8, and the hypotenuse is always across the right angle.

15/17 would be sine instead of cosine

8 0

3 years ago

Graph the polygon and its image after a dilation centered at C(1,1)

Grace [21]

Now we proceed to present the graphs of the polygon and its image with the help of a graphing tool. The original polygon is highlighted in red and the image is highlighted in blue.

DIlations are defined by the follwing operation:

$P'(x,y) = O(x,y) + k\cdot [P(x,y)-O(x,y)]$ (1)

Where:

$O(x,y)$ - Center of dilation.
$k$ - Dilation factor.
$P(x,y)$ - Original point of the polygon.
$P'(x,y)$ - Dilated point of the polygon.

If we know that $O(x,y) = (1,1)$ , $k = 0.5$ , $J(x,y) = (3,1)$ , $K(x,y) = (5, -3)$ , $L(x,y) = (5,5)$ and $M(x,y) = (3,7)$ , then the coordinates of the new polygon are:

$J'(x,y) = (1,1) + 0.5\cdot [(3,1)-(1,1)]$

$J'(x,y) = (1,1)+0.5\cdot (2,0)$

$J'(x,y) = (1,1) +(1,0)$

$J'(x,y) = (2,1)$

$K'(x,y) = (1,1) + 0.5\cdot [(5,-3)-(1,1)]$

$K'(x,y) = (1,1) +0.5\cdot (4,-4)$

$K'(x,y) = (1,1) +(2,-2)$

$K'(x,y) = (3,-1)$

$L'(x,y) = (1,1) + 0.5\cdot [(5,5)-(1,1)]$

$L'(x,y) = (1,1) + 0.5\cdot (4,4)$

$L'(x,y) = (1,1) + (2,2)$

$L'(x,y) = (3,3)$

$M'(x,y) = (1,1) + 0.5\cdot [(3,7)-(1,1)]$

$M'(x,y) = (1,1) +0.5\cdot (2,6)$

$M'(x,y) = (1,1) +(1,3)$

$M'(x,y) = (2,4)$

Now we proceed to present the graphs of the polygon and its image with the help of a graphing tool. The original polygon is highlighted in red and the image is highlighted in blue.

We kindly invite to check this question on dilations: brainly.com/question/13176891