Answer:
Step-by-step explanation:
To prove Δ ABC similar to ΔDBE we can consider
Segments AC and DE are parallel.
⇒ DE intersects AB and BC in same ratio.
AB is a transversal line passing AC and DE.
⇒∠BAC=∠BDE [corresponding angles]
Angle B is congruent to itself due to the reflexive property.
All of them are telling a relation of parts of ΔABC to ΔDBE.
The only option which is not used to prove that ΔABC is similar to ΔDBE is the first option ,"The sum of angles A and B are supplementary to angle C".
Step 1: -3(x+2y=3)
Step 2: -3x-6y=-9
Step 3: 3x cancels out with -3x
Step 4: add 2y and -2y, and add 3 and 5
Step 5: the answer is NO SOLUTION or INFINITE SOLUTION
I honestly don’t know man but don’t worry, if you get a bad score it won’t matter in 30 years :)
Answer:
x ≈ 1.32, x ≈ - 5.32
Step-by-step explanation:
Given
x² + 4x - 7 = 0 ( add 7 to both sides )
x² + 4x = 7
To complete the square
add ( half the coefficient of the x- term )² to both sides
x² + 2(2)x + 4 = 7 + 4
(x + 2)² = 11 ( take the square root of both sides )
x + 2 = ± 
( subtract 2 from both sides )
x = - 2 ± 
Thus
x = - 2 - ≈ - 5.32 ( to 2 dec. places )
x = - 2 + ≈ 1.32 ( to 2 dec. places )
