Y=-2+-4x m=slope b=y-intercept
m=-4
b=-2
Answer: x=10 and y=25
Step-by-step explanation:
ok, so since we know straight angles=180 degrees, so since one part=100, the other smaller angle=80. This means that 11x-30=80 and 5y-25=100. 11*10=110, and 110-30=80, so x=10. 5*25=125, and 125-25=100, so y=25. And lol. me too. When I used to do these problems, I was stuck for a very, very long time. Just try to use logic most of the time.
This is a little long, but it gets you there.
- ΔEBH ≅ ΔEBC . . . . HA theorem
- EH ≅ EC . . . . . . . . . CPCTC
- ∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
- ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
- ∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
- ∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
- ΔDAC ≅ ΔDAG . . . HA theorem
- DC ≅ DG . . . . . . . . . CPCTC
- ∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
- ∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
- ∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
- ∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
- (∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
- ∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
- This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)
Slope would be undefined because there are many many y-values for one x value. This is impossible for ANY function. To be specific, the correct equation that names the graph is <em>x=4</em>. Please note that this is <u>not</u> a function. If it goes straight up, it is <u>UNDEFINED</u>!!!
Hope I helped!
The answer is <u><em>D</em></u>
Answer:
GCF: 2
Step-by-step explanation:
GCF: 2
(2*16) + (2*27)
