32. (a) For an even function, f(x) = f(-x). Given f(5) = 3, we know f(-5) = 3.
Therefore (-5, 3) is also on the graph.
For an odd function, f(-x) = -f(x). Given f(5) = 3, we know f(-5) = -3.
Therefore (-5, -3) is also on the graph.
33. f(-x) = -f(x). The function is odd.
34. f(-x) = x/(x-1) ≠ -f(x) ≠ f(x). The function is neither even nor odd.
35. f(-x) = f(x). The function is even.
So you’d add 5 and 3 so it’s 8 ft/s
You then divide 220/8
So it should be 27.5 seconds.
Sorry about that other guy. Hope this helps!
Answer:
- A-B=7x-3x+10-4x²-66+4
- 4x-52-4x²
hope it helps
stay safe healthy and happy.
Answer:
Option (1) will be the answer.
Step-by-step explanation:
Coordinates of the points A and B lying on the line f are (0, 2) and (2, 0) respectively.
Slope of the line f,
After dilation of line f by a scale factor of 2, coordinates of A' and B' will be,
Rule for dilation,
(x, y) → (kx, ky)
Where k = scale factor
A(0, 2) → A'(0, 4)
B(2, 0) → B'(4, 0)
Slope of line f',
Since, 
Therefore, both the lines f and f' will be parallel.
Option (1) will be the answer.
The minimum value of a function is the place where the graph has a vertex at its lowest point.
There are two methods for determining the minimum value of a quadratic equation. Each of them can be useful in determining the minimum.
(1) By plotting graph
We can find the minimum value visually by graphing the equation and finding the minimum point on the graph. The y-value of the vertex of the graph will be the minimum.
(2) By solving equation
The second way to find the minimum value comes when we have the equation y = ax² + bx + c.
If our equation is in the form y = ax^2 + bx + c, you can find the minimum by using the equation min = c - b²/4a.
The first step is to determine whether your equation gives a maximum or minimum. This can be done by looking at the x² term.
If this term is positive, the vertex point will be a minimum; if it is negative, the vertex will be a maximum.
After determining that we actually will have a minimum point, use the equation to find it.
