<em>AC bisects ∠BAD, => ∠BAC=∠CAD ..... (1)</em>
<em>thus in ΔABC and ΔADC, ∠ABC=∠ADC (given), </em>
<em> ∠BAC=∠CAD [from (1)],</em>
<em>AC (opposite side side of ∠ABC) = AC (opposite side side of ∠ADC), the common side between ΔABC and ΔADC</em>
<em>Hence, by AAS axiom, ΔABC ≅ ΔADC,</em>
<em>Therefore, BC (opposite side side of ∠BAC) = DC (opposite side side of ∠CAD), since (1)</em>
<em />
Hence, BC=DC proved.
8 ÷ (7 - 9) * ( 4 + (-4) ) <--- notice that bolded part.
4 + (-4) = 4 - 4 = 0.
after that, you're pretty much dividing and multiplying by 0, so
whatever ÷ whatever * whatever * 0 = 0.
53.67 to one decimal place is 53.7 since you round the .6 to the .7
Answer:
294 - 14 = 280 ÷ 5 = 56
In conclusion, there were 56 Students in each Bus.
