Answer:
The image of OH is ON.
Step-by-step explanation:
Te figure GIKMPR is a regular hexagon. The number of vertices of a regular hexagon is 6. The central angle between any two consecutive vertices is 60 degree.
The dashed line segments form 30 degree angles.
If we rotate the hexagon 180 degree about O, then the each point shifts to 6th place from its original place.
Since we rotate the hexagon 180 degree about O, so the image and preimage lies on a straight line. Because a straight line make angle of 180 degree.
The line OH and ON lies on a straight line therefore the image of OH is ON.
I believe it’s 12. But I’m not sure
Answer:
y - value of the vertex is 49.
Step-by-step explanation:
Given function is f(x) = -(x - 3)(x + 11)
f(x) = -(x² - 3x + 11x - 33)
= -(x² + 8x - 33)
= -(x² + 8x + 16 - 49)
= -[(x + 4)² - 49]
= -(x + 4)² + 49
Comparing this equation with the vertex form of a quadratic function,
f(x) = -(x - h)² + k
Where (h, k) is the vertex of the function.
Vertex of the parabola is (-4, 49)
Therefore, y-value of the vertex is 49.
Answer:
a) integral = 24.72
b) |Error| ≤ 0.4267
Step-by-step explanation:
a)
The integral:
can be approximated with the midpoint rule, as follows:
6.7*(0.8 - 0.0) + 8.9*(1.6 - 0.8) + 6.9*(2.4-1.6) + 8.4*(3.2 - 2.4) = 24.72
(that is, all the intervals are 0.8 units length and f(x) is evaluated in the midpoint of the interval)
b)The error bound for the midpoint rule with <em>n</em> points is:
|Error| ≤ K*(b - a)^3/(24*n^2)
where <em>b</em> and are the limits of integration of the integral and K = max |f''(x)|
Given that -5 ≤ f''(x) ≤ 1, then K = 5. Replacing into the equation:
|Error| ≤ 5*(3.2 - 0)^3/(24*4^2) = 0.4267
