The coordinate grid shows points A through K. What point is a solution to the system of inequalities?

y < −2x + 10
y < 1 over 2x − 2

coordinate grid with plotted ordered pairs, point A at negative 5, 4 point B at 4, 7 point C at negative 2, 7 point D at negative 7, 1 point E at 4, negative 2 point F at 1, negative 6 point G at negative 3, negative 10 point H at negative 4, negative 4 point I at 9, 3 point J at 7, negative 4 and point K at 2, 3

I
B
A
F

hi,

6x times 4 is 24x and 5 times 4 is 20, so the answer is 24x+20.

hope this and for a little humor i am groot.

i'm sure goog can you get that answer

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.