Answer:
Yes, they are parallel
Step-by-step explanation:
Find the slopes of the lines using rise over run (y2 - y1) / (x2 - x1)
Find the slope of line M:
(y2 - y1) / (x2 - x1)
(2 - 0) / (-3 + 5)
2/2
= 1
Find the slope of line N:
(y2 - y1) / (x2 - x1)
(4 - 2) / (0 + 2)
2/2
= 1
So, the two lines have the same slope of 1, meaning that they are parallel because parallel lines have the same slope.
They are parallel.
Given:
Point A with coordinates: (4, 2) is in the first quadrant.
The point is translated to the rule (x, y) --> (x+1, y-5)
Now, point A becomes (4+1, 2-5) = (5, -3) which is now in quadrant 4.
The point is reflected across the y-axis. Now, it is in the third quadrant.
If <em>x</em> + 1 is a factor of <em>p(x)</em> = <em>x</em>³ + <em>k</em> <em>x</em>² + <em>x</em> + 6, then by the remainder theorem, we have
<em>p</em> (-1) = (-1)³ + <em>k</em> (-1)² + (-1) + 6 = 0 → <em>k</em> = -4
So we have
<em>p(x)</em> = <em>x</em>³ - 4<em>x</em>² + <em>x</em> + 6
Dividing <em>p(x)</em> by <em>x</em> + 1 (using whatever method you prefer) gives
<em>p(x)</em> / (<em>x</em> + 1) = <em>x</em>² - 5<em>x</em> + 6
Synthetic division, for instance, might go like this:
-1 | 1 -4 1 6
... | -1 5 -6
----------------------------
... | 1 -5 6 0
Next, we have
<em>x</em>² - 5<em>x</em> + 6 = (<em>x</em> - 3) (<em>x</em> - 2)
so that, in addition to <em>x</em> = -1, the other two zeros of <em>p(x)</em> are <em>x</em> = 3 and <em>x</em> = 2
Start by using trig to find the length of the line LJ
The triangle KJL (big right angled triangle) has been given the following dimensions
Hypotenuse =

The adjacent angle is 30 degrees
Since we have the hypotenuse and the angle we must use the equation
opposite = Sin(angle) x Hypotenuse
Opposite= sin30 x

Opposite=

Therefore line LJ is

Now look at the smaller right angled triangle (LMJ)
Hypotenuse is the line LJ which is

The adjacent angle is 45
Since we have hypotenuse and angle we must use the equation opposite = sin(angle) * h
therefore
x=

* sin45= 4