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:
t = 0
Step-by-step explanation:
(7t - 2) - (-3t + 1) = -3(1 - 3t)
because you can not do anything inside the parantheses, you distribute. there aren't any numbers on the left side of the parantheses on the left side of the equation so we just imagine the number 1. on the right side of the equation, just distribute normally
[ 1(7t - 2) -1(-3t + 1) = -3(1 - 3t) ]
will be
7t - 2 + 3t - 1 = -3 + 9t
add like terms
10t - 3 = -3 + 9t
subtract 9t from both sides
t - 3 = -3
add 3 to both sides
t = 0
Answer:
A) 24
Step-by-step explanation:
Total weight - loaded weight = space remaining in the trailer: 1050kg - 82kg = 968kg
space remaining ÷ the number of transport: 968 ÷ 40 = 24.2 OR 24
Hope it helps you
Answer:
24x^2+112x+120
Step-by-step explanation:
distribute the 2
(6x+10)(4x+12)
foil it
24x^2+112x+120
