Answer:
x = 41
Step-by-step explanation:
x-23=18
Add 23 to each side
x-23+23=18+23
x =41
When x = -1, y = - 3 ((-1)^2-4)
When x = 0, y = -4 ((0)^2-4)
When x = 5, y = 21 ((5)^2-4)
When x = 13 y = 165 ((13)^2-4)
You just substitute the values of X into the equation
Answer:
draw 4
Step-by-step explanation:
uno draw 4 color red I WIN!
JK IT'S 14.7u^8o
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).)
Answer:
1/4 for both
Step-by-step explanation:
2/8 -- you divide 2 by 2 you get 1, 8 divided by 2 is 4 -- 1/4
3/12 -- you divide 3 by 3 you get 1, 12 divided by 3 is 4 -- 1/4
** tip: what you do to one half of the fraction you have to do to the other
hope this helps!