On a coordinate plane, a square and a point are shown. The square has points R prime (negative 8, 1), S prime (negative 4, 1), T prime (negative 4, negative 3), and U prime (negative 8, negative 3). Point S is at (3, negative 5). Square RSTU is translated to form R'S'T'U', which has vertices R'(–8, 1), S'(–4, 1), T'(–4, –3), and U'(–8, –3). If point S has coordinates of (3, –5), which point lies on a side of the pre-image, square RSTU? (–5, –3) (3, –3) (–1, –6) (4, –9)

(–1, –6)

Step-by-step explanation:

Given that S(3, –5) was translated to S'(–4, 1), the transformation rule is (x-7, y+6). Then, the coordinates of square RSTU are:

R'(–8, 1) -> (-8+7, 1-6) -> R(-1, -5)

S'(–4, 1) -> (-4+7, 1-6) -> S(3, -5)

T'(–4, –3) -> (-4+7, -3-6) -> T(3, -9)

U'(–8, –3) -> (-8+7, -3-6) -> U(-1, -9)

The point (–1, –6) lies on the segment RU

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it

according to egenuity the answer is

step-by-step explanation:

d. 344.22