Part I)
The module of vector AB is given by:
lABl = root ((- 3) ^ 2 + (4) ^ 2)
lABl = root (9 + 16)
lABl = root (25)
lABl = 5
Part (ii)
The module of the EF vector is given by:
lEFl = root ((5) ^ 2 + (e) ^ 2)
We have to:
lEFl = 3lABl
Thus:
root ((5) ^ 2 + (e) ^ 2) = 3 * (5)
root ((5) ^ 2 + (e) ^ 2) = 15
Clearing e have:
(5) ^ 2 + (e) ^ 2 = 15 ^ 2
(e) ^ 2 = 15 ^ 2 - 5 ^ 2
e = root (200)
e = root (2 * 100)
e = 10 * root (2)
Answer: The length of segment GT is 6
Step-by-step explanation: Have a nice day!✌️
Quadratic f(x) = (x -h)² +k has vertex (h, k) and axis of symmetry x=h. When k is negative, the number of real solutions is 2, because both branches of the function cross the x-axis.
In your equation, h = -3 and k = -8.
Axis of symmetry: x = -3
Vertex: (-3, -8)
Number of real solutions: 2
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