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Given that ∠B ≅ ∠C.
to prove that the sides AB = AC
This can be done by the method of contradiction.
If possible let AB =AC
Then either AB>AC or AB<AC
Case i: If AB>AC, then by triangle axiom, Angle C > angle B.
But since angle C = angle B, we get AB cannot be greater than AC
Case ii: If AB<AC, then by triangle axiom, Angle C < angle B.
But since angle C = angle B, we get AB cannot be less than AC
Conclusion:
Since AB cannot be greater than AC nor less than AC, we have only one possibility. that is AB =AC
Hence if angle B = angle C it follows that
AB = AC, and AB ≅ AC.
Answer:
Step-by-step explanation:
In order to determine the information you're being asked for, you need to complete the square on that quadratic. The first step is to move the constant over to the other side of the equals sign:
Here would be the step where, if the leading coefficient isn't a 1, you'd factor it out. But ours is a 1, so we're good there. Now take half the linear term (the term with the single x on it), square it, and add it to both sides. Our linear term is a -2. Half of -2 is -1, and -1 squared is +1. We add +1 to both sides giving us this:
Now we'll clean it up a bit. The right side becomes a 4, and the left side is written as its perfect square binomial, which is the whole reason we did this. That binomial is
(set equal to the 4 here). Now we'll move the 4 back over and set the whole thing back equal to y:
From this it's apparent what the vertex is: (1, -4),
the axis of symmetry is x = 1, and
the y-intercept is found by setting the x's equal to 0 in the original equation and solving for y. So the y-intercept is (0, -3).
Your choice for the correct answer is the very last one there.
Answer:
2362 ...........................
