Triangle rst has vertices r(2, 0), s(4, 0), and t(1, –3). the image of triangle rst after a rotation has vertices r'(0, –2), s'(0, –4), and t'(–3, –1). which rule describes the transformation? \

Answer with explanation:

Pre -image= Vertices of Δ R ST=R(2, 0), S(4, 0), and T(1, –3)

Image of Δ R ST after rotation= R'(0, –2), S'(0, –4), and T'(–3, –1)

Pre-Image lies in Fourth Quadrant and Image lies in Third Quadrant.

If triangle is rotated by different angles in anticlockwise direction,then

If triangle is rotated by different angles in Clockwise direction,then

⇒→So, Pre image that is ,Δ R ST having vertices ,R(2, 0), S(4, 0), and T(1, –3) when rotated by either 90° in clockwise direction or by 270°, in anticlockwise Direction to get Image Δ R' S'T' having vertices R'(0, –2), S'(0, –4), and T'(–3, –1)  .

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15