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:
(a) x > 4 (b) y < -2
Step-by-step explanation:
Domain is referring to the x-values while the range is referring to the y-values.
Since the function (the line) has a circle at the point (4, -2), the function will be exclusive at that coordinate.
The line goes to infinity for the x-values from 4, so you write x > 4 or ∞ > x > 4.
Similarly, the line goes to infinity for the y-values from -2, so you write y < -2 or -∞ < y < -2.
Answer:
y = 11 1/2
Step-by-step explanation:
7y - 63 = y + 3
-y. -7
6y - 63 = 3
+ 63. + 63
6y = 69
/6. /6
y = 11 1/2
Answer is A
if x+y=5
y= -x+5
2x + (-x+5) =-2
2x -x +5 = -2
x+5 = -2
x=-7
