Answer:
Step-by-step explanation:
<u>Given</u>
- m∠ABC = x°
- m∠BCD = 25°
- m∠CDE = 55°
- m∠DEF = 3x°
Add two more parallel lines passing through points C and D.
Consider alternate interior angles formed by the four parallel lines.
<u>The angles between the two middle lines are equal to:</u>
- m∠BCD - m∠ABC = m∠CDE - m∠DEF
<u>Substitute values and solve for x:</u>
- 25 - x = 55 - 3x
- 3x - x = 55 - 25
- 2x = 30
- x = 15
m∠ABC = 15°
N - the number
6 × n - 11 = 6n - 11
Did somebody say you're supposed to draw the graph of the equation ?
Is that the assignment ?
OK. Just like every other equation you need to graph, get it in the
standard form, where 'y' is all alone on one side, and everything else
is on the other side. When you do that, you'll be able to spot the slope
and y-intercept of the line, or get some points, or whatever you want.
4y + 12 = 0
Subtract 12 from each side: 4y = -12
Divide each side by 4: y = -3
There's the equation you can handle.
The y-intercept is -3, and the slope is zero.
Would you like some points ? OK. Pick a couple of values for 'x',
and calculate the value of 'y' for each one:
The first value I picked for 'x': x = 72
The equation is y=-3, so when x=72, y=-3. The point is (72, -3)
The second value I picked for 'x' is: x = 1
The equation is y=-3, so when x=1, y=-3. The second point is (1, -3).
The third value I picked for 'x' is 4 billion.
The equation is y=-3, so when x=4 billion, y=-3. The third point is (1, -3).
Do you see what's going on here ? Your original equation didn't even
have 'x' in it, so we could tell right away that when the graph is drawn,
the value of 'y' at every point can't depend on 'x'.
When we simplified the equation and got it in standard form, we found that
the slope of the graph is zero. That means the graph doesn't rise or fall ...
it's just a horizontal line. Sure enough, the height of points on the line
doesn't depend on 'x'. The value of 'y' at every point on the line is -3 .
Answer:
x = 11º
Step-by-step explanation:
1. Notice Parallel Lines
2. Understand Angle Relationships When Parallel Lines Are Present (e.g., alternate interior/exterior)
3. ∠CAB ≅ ∠DCA ∴ m∠CAB = 33º
4. Use exterior angle theorem: the sum of non-adjacent angles of the same triangle which the exterior angle is drawn is equal to the measure of that angle.
5. Therefore write and solve the equation 2x + 33º (sum of non-adjacent interior angles) = 5x (exterior angle).
- 2x + 33 = 5x (C.L.T or <u>C</u>ombine <u>L</u>ike <u>T</u>erms)
- 3x = 33 (inverse operations; divide by 3)
- x = 11º (remember to apply units)
