The answer is there I explained the process.
Try grouping in paires
4x^3 + 28x^2 + 7x + 49
= 4x^2(x + 7) + 7(x + 7) - note that (x + 7) is common to the 2 parts, so
= (4x^2 + 7)(x + 7)
When two parallel lines are intersected by a transversal, the same-side exterior angles are supplementary. That means that their sum is 180.
Using that logic, if the two roads were parallel, then the sum of their same-side exterior angles will add up to 180. Yet their same-side exterior angles add up to 170 (130 + 40 = 170), hence they can't be parallel.
See the drawing attached below.
Using supplmenatry angles (two angles whose sum of measures add up to 180 or a straight line), we can say that:
m<DIE + m<HID = 18
40 + m<HID = 180
m<HID = 140
Similarly:
m<BHC + m<CHI = 180
130 + m<CHI = 180
m<CHI = 50
Using verticle angles therome, (when two lines intersect, the angles opposite to eachother are congruent, or have the same measure), we can say that:
m<DIE = m<GIH = 40
m<GIE = m<HID = 140
m<CHI = m<AHB = 50
m<BHC = m<AHI = 130
The rhombus has equal side lengths while the parallelogram doesn't have equal side lengths.
Answer:
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