On a coordinate plane, a solid straight line has a positive slope and goes through (negative 3, negative 5) and (0, negative 4). everything to the right of the line is shaded.

which linear inequality is represented by the graph?

y ≥ one-thirdx – 4
y ≤ one-thirdx – 4
y ≤ one-thirdx + 4
y ≥ one-thirdx + 4

y ≤ One-thirdx – 4

Step-by-step explanation:

The key hint to know this answer is to understand that the line cuts the y-axis at -4. And, we can know that y is less than or equal to (≤) 1/3x-4 for the shaded area to the RIGHT of the line, telling us that y exists for all values from -4 and under.

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl

shift 4 units up, and 1 unit left.

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