Answer: The required length of the segment AA' is 11 units.
Step-by-step explanation: Given that the point A(5, 11) is reflected across the X-axis.
We are to find the length of the segment AA'.
We know that
if a point (x, y) is reflected across X-axis, then its co-ordinates becomes (x, -y).
So, after reflection, the co-ordinates of the point A(5, 11) becomes A'(5, -11).
Now, we have the following distance formula :
The DISTANCE between two points P(a, b) and Q(c, d) gives the length of the segment PQ as follows :
Therefore, the length of the segment AA' is given by
Thus, the required length of the segment AA' is 11 units.
(-2,5)(9,-6)
slope(m) = (-6-5) / (9 - (-2) = -11/11 = -1
y = mx + b
slope(m) = -1
(-2,5)...x = -2 and y = 5
now we sub and solve for b, the y int
5 = -1(-2) + b
5 = 2 + b
5 - 2 = b
3 = b
so ur equation in slope intercept form is : y = -x + 3
Answer:
7 and 1
Step-by-step explanation:
√50 is the sum of the square of two legs
The answer is likely to be 7 and 1 (because 7^2 + 1^1 = 50)
Answer:
7x - 5 7x - 5
(f/g) = ----------------- = ----------------
2x - 4 2(x - 2)
Step-by-step explanation:
Given f(x) = 7x-5 and g(x)=2x-4, we form the quotient f/g as follows:
7x - 5 7x - 5
(f/g) = ----------------- = ----------------
2x - 4 2(x - 2)
Note that this is not defined for x - 2, due to division by zero.
