Answer:
(the second option)
Step-by-step explanation:
Hope this helps!
Write a direct variation equation that relates x
1 answer:
You might be interested in
Answer:
Answer:
Step-by-step explanation:
The domain of all polynomials is all real numbers. To find the range, let's solve that quadratic for its vertex. We will do this by completing the square. To begin, set the quadratic equal to 0 and then move the -10 over by addition. The first rule is that the leading coefficient has to be a 1; ours is a 2 so we factor it out. That gives us:
The second rule is to take half the linear term, square it, and add it to both sides. Our linear term is 2 (from the -2x). Half of 2 is 1, and 1 squared is 1. So we add 1 into the parenthesis on the left. BUT we cannot ignore the 2 sitting out front of the parenthesis. It is a multiplier. That means that we didn't just add in a 1, we added in a 2 * 1 = 2. So we add 2 to the right as well, giving us now:
The reason we complete the square (other than as a means of factoring) is to get a quadratic into vertex form. Completing the square gives us a perfect square binomial on the left.
and on the right we will just add 10 and 2:
Now we move the 12 back over by subtracting and set the quadratic back to equal y:
From this vertex form we can see that the vertex of the parabola sits at (1,-12). This tells us that the absolute lowest point of the parabola (since it is positive it opens upwards) is -12. Therefore, the range is R={y|y ≥ -12}
Answer:
Step-by-step explanation:
<u>Similar Triangles</u>
Lines ST and QR are parallel. Thus, angles S and Q are congruent, and angles T and R are congruent.
Considering the triangle PQR, the sum of its internal angles must be 180°:
Substituting the known values:
Solving for R:
Angles S and Q are congruent, thus
Angles T and R are congruent, thus
Summarizing: