Please help select the algebraic rule that describes the dialtion of ABC to A'B'C

A.(3x,3y)
B.(2x,2y)
C.(1/3x,1/3y)

A.(3x,3y)

Step-by-step explanation:

In the given graph,

Triangle A'B'C' is an image of Triangle ABC after dilation.

Algebraic rule for dilation (with center =origin)

    , where k= scale factor.

Coordiantes of A = (1,1)  Coordinates of A' = (3,3)

Scale factor =

Hence, the required algebraic rule : (3x,3y)

The correct option is A.

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

your answer is the frist one

hop i