The figure shows triangle abc with medians a f, bd, and ce. segment a f is extended to h in such a way that segment gh is congruent to segment ag. which conclusion can be made based on the given conditions? segment gf is parallel to segment eb. segment eg is parallel to segment bh. segment bh is congruent to segment hc. segment eg is congruent to segment gd. only answer if you confidently know, .

From given Conditions we conclude that is parallel to

Step-by-step explanation:

Given: , and are medians of ΔABC.

Option 1:

is not parallel to because on extending both segment they are intersecting at A but parallel lines never intersects.

Option 2:

is parallel to because E is mid point of and G is mid point of then according to Mid Point Theorem is Parallel to .

Option 3:

is not congruent to . It is clear from figure.

Option 4:

is not congruent to . Since from above option is not congruent to then using Mid Point theorem in ΔABH and ΔAHC.

We get,

In ΔABH is parallel to & ....... Eqn (1)

Similarly,

In ΔAHC

........ Eqn (2)

So, from eqn (1) & (2)

≠

left 3, up 4

step-by-step explanation:

just count the values of x & y of the vertex of the graph and see how many units in the x and y direction will get you from the f(x) graph to the f(x) graph

shift 4 units up, and 1 unit left.

it will translate f(x) = 3^x to g(x) = 3^(x+1) + 4