The best way to solve this problem is to graph all variables, velocity being the dependent variable (y-axis) and time being the independent variable (x-axis) . Since we are given two points for the two variables, we use the point-slope formula to get the slope which happens to be -1. We then use the standard form y = mx + b and substitute m with -1. Next we want to find b, we will again use the point slope formula now knowing m to be -1. we chose point 0 for time(x-axis) and at zero, we obtain a value of your b which happens to be +19. A line graph would better visualize my point. Cheers!
Answer:
Step-by-step explanation:
a. The vertex is (1, -9). False. The vertex is (-1, -9).
b. The graph opens upward. True. That coefficient '6' is positive.
c. The graph is obtained by shifting the graph of f(x) = 6(x + 1)2 up 9 units. False. It's DOWN 9 units.
d. The graph is steeper than the graph of f(x) = x^2. True. The given function increases 6 times faster than does x^2.
e. The graph is the same as the graph of f(x) = 6x2 + 12x - 3.
Let's put f(x) = 6(x + 1)^2 -9 into standard form. Expanding the first term, we get f(x) = 6(x^2 + 2x + 1) - 9, or f(x) = 6x^2 + 12x + 6 - 9. YES, TRUE
I think the answer is 16x^2+1
Answer:
The correct option is O B'
Step-by-step explanation:
We have a quadrilateral with vertices A, B, C and D. A line of reflection is drawn so that A is 6 units away from the line, B is 4 units away from the line, C is 7 units away from the line and D is 9 units away from the line.
Now we perform the reflection and we obtain a new quadrilateral A'B'C'D'.
In order to apply the reflection to the original quadrilateral ABCD, we perform the reflection to all of its points, particularly to its vertices.
Wherever we have a point X and a line of reflection L and we perform the reflection, the new point X' will keep its original distance from the line of reflection (this is an important concept in order to understand the exercise).
I will attach a drawing with an example.
Finally, we only have to look at the vertices and its original distances to answer the question.
The vertice that is closest to the line of reflection is B (the distance is 4 units). We answer O B'
