Answer:
Option B is correct.
Step-by-step explanation:
We have given a triangle ABC and EDC please look at the figure
We can see that AE and BD are transversal therefore, ∠EAB=∠AED being alternate interior angles
And ∠ACB=∠DCE are vertically opposite angles hence, equal
So, by AA similarity postulate the above to triangles are similar
ΔABC ΔEDC
Therefore, Option B is correct that is Triangle ABC is similar to triangle EDC , because m∠3 = m∠4 and m∠1 = m∠5
NOTE: m∠3 = m∠4 corresponds to m∠ACB=m∠DCE
And m∠1 = m∠5 corresponds to m∠EAB=m∠AED
Answer:
Step-by-step explanation:
Use FOIL method
1) (x +4 )(3x +2) = x*3x + x*2 + 4*3x + 4*2
= 3x² + 2x + 12x + 8
= 3x² + 14x + 8
4) (4x - 1) (2x + 7) = 4x*2x + 4x*7 + (-1)*2x + (-1) * 7
= 8x² + 28x - 2x - 7
= 8x² + 26x - 7
