The <em>quadratic</em> function g(x) = (x - 5)² + 1 passes through the points (2, 10) and (8, 10) and has a vertex at (5, 1).
<h3>How to analyze quadratic equations</h3>
In this question we have a graph of a <em>quadratic</em> equation translated to another place of a <em>Cartesian</em> plane, whose form coincides with the <em>vertex</em> form of the equation of the parabola, whose form is:
g(x) = C · (x - h)² - k (1)
Where:
- (h, k) - Vertex coordinates
- C - Vertex constant
By direct comparison we notice that (h, k) = (5, 1) and C = 1. Now we proceed to check if the points (x, y) = (2, 10) and (x, y) = (8, 10) belong to the parabola.
x = 2
g(2) = (2 - 5)² + 1
g(2) = 10
x = 8
g(8) = (8 - 5)² + 1
g(8) = 10
The <em>quadratic</em> function g(x) = (x - 5)² + 1 passes through the points (2, 10) and (8, 10) and has a vertex at (5, 1).
To learn more on parabolae: brainly.com/question/21685473
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Answer:
Step-by-step explanation:
The length of the midsegment is half of the base in this case.
12x = 8x + 40
4x=40
x=10
So it's 60
P(3) = 1/8
P(5)=1/8
Events are independent.
P(3, then 5) = 1/8 *1/8 =1/16
Answer is 1/16.
Check the slopes
the first 2 points have slope (3-1) / (5-1) = 2/4 = 1/2
points 2 and 3 slope = 4/2 = 2
3 and 4:- slope = -2/-4 = 1/2
4 and 1:- slope = 4/2 = 2
so we have 2 pairs of opposite sides which are parallel
looks like a parallelogram or a rhombus
if adjacent sides are equal its a rhombus
length betwee pts 1 and 2 = sqrt(2^2 + 4^2) = sqrt20
.. ... ... .. .. .. ...2 and 3 = sqrt(4^2 + 2^2) = sqrt20
so it is a rhombus