11. 10ac × 6ab × (–2bc) = ? A. 14a2b2c2 B. 2a + 2b + 2c + 14 C. –120a2b2c2 D. 2a + 2b +2c – 120

Accepted Solution

Answer:C. –120a2b2c2Step-by-step explanation:We'll just the multiplications one at a time...10ac × 6ab × (–2bc) = 60a²bc * (-2bc)  (multiplying 10ac × 6ab) 60a²bc * (-2bc)  = -120a²b²c²We just have to multiply the numbers together (like 10 and 6), then all similar letters together (like ac * ab = a²bc).  If a letter is already present (like a in this  example), then we add up its powers a * a = a², a² * a would be a³ and so on).  If the letter is not present (like 'b' in 'ac', we suppose it's there as b^0 which is equal to 1)... so we still add up its exponent (power)... so it makes b^0 * b = b