F(x)=(a*x) - (b*4), f(0)=16, and f(3)=25. what is a+b?

This is a quite interesting problem.

Since it is normally unsolvable without the functions, the answer must lay somewhere in the examples given.

In this case, it lays in f(0).

Why would that be? Because anything times 0 is equal to 0.

In f(0), x is equal to 0, so a*x is therefore equal to 0.

Knowing that, the only thing that is left behind is, when you substitute in 0 for x:

16 = -(b*4)

Therefore, we can then solve to get:

-16 = b * 4

-4 = b

And now it's fairly simple to get a: just plug b into f(3) and solve.

25 = (3a) - (-4*4)

25 = 3a + 16

25 - 16 = 3a

9 = 3a

a = 3

And now to solve for what the question is looking for, just add a and b.

3 + -4 = -1

And there's your answer.

25

step-by-step explanation:

∠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.