∠BDC and ∠AED are right angles, is a piece of additional information is appropriate to prove △ CEA ~ △ CDB
Triangle AEC is shown. Line segment B, D is drawn near point C to form triangle BDC.
<h3> What are Similar triangles?</h3>
Similar triangles, are those triangles which have similar properties,i.e. angles and proportionality of sides.
Image is attached below,
as shown in figure
∡ACE = ∡BCD ( common angle )
∡AED = ∡BDC ( since AE and BD are perpendicular to same line EC and make right angles as E and C)
∡EAC =- ∡DBC ( corresponding angles because AE and BD are parallel lines)
Thus, △CEA ~ △CDB , because of the two perpendiculars AE and BD.
Learn more about similar triangles here:
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If you need to simplify the equation it would be:
+2r-2n+7
If not sorry..
Hope This Helps!
Answer:
y = (2 x)/3 + 1/4
Step-by-step explanation:
Solve for y:
8 x - 12 y = -3
Subtract 8 x from both sides:
-12 y = -8 x - 3
Divide both sides by -12:
Answer: y = (2 x)/3 + 1/4
You need to plot each line and see where they intersect.
Line 1: y = x+2
Plot the y-intercept (0,2) because of the +2 in the equation.
From (0,2), count "up 1, right 1" to get a second point, because the slope is 1.
From that new point, repeat the "up 1, right 1" to plot a third point.
Connect the dots to make your line.
Line 2: y = -1/3 x - 2
Repeat the same process, using the the y-intercept and slope for this line.
Then identify where they intersect.
- 30m^2 - 10m + 30
distribute your negative 10, you will first get 30m, then - 30m^2.
next distribute the negative 5, you will get - 40m, then 30.
combine like terms, and make sure its in standard form, - 30m^2 - 10m + 30
