Question

एक AABC, DE समानान्तर है BC के तथा AD = (x+3), DB (3x+19), AE
(3x+4), तो x का मान ज्ञात करो।
10 find the probability​

Answer 1

Answer:

$3x + 19 \\ \\ 3x = 19 \\ \\ x = \frac{19}{3} \\ \\ now \\ \\ 3x + 4 = 0 \\ \\ 3x = 4 \\ \\ x = \frac{4}{3} answer$