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For 
t!=0, which of the following expressions is equivalent to 
(81t^(2)+27 t)/(9t^(2)+9t) ?
Choose 1 answer:
(A) 
(9t+27)/(t+9)
(B) 
(9t+3t)/(t+1t)
(c) 
(9t+3)/(t+1)
(D) 
(9t+9)/(t+3)

For t0 t \neq 0 , which of the following expressions is equivalent to 81t2+27t9t2+9t \frac{81 t^{2}+27 t}{9 t^{2}+9 t} ?\newlineChoose 11 answer:\newline(A) 9t+27t+9 \frac{9 t+27}{t+9} \newline(B) 9t+3tt+1t \frac{9 t+3 t}{t+1 t} \newline(C) 9t+3t+1 \frac{9 t+3}{t+1} \newline(D) 9t+9t+3 \frac{9 t+9}{t+3}

Full solution

Q. For t0 t \neq 0 , which of the following expressions is equivalent to 81t2+27t9t2+9t \frac{81 t^{2}+27 t}{9 t^{2}+9 t} ?\newlineChoose 11 answer:\newline(A) 9t+27t+9 \frac{9 t+27}{t+9} \newline(B) 9t+3tt+1t \frac{9 t+3 t}{t+1 t} \newline(C) 9t+3t+1 \frac{9 t+3}{t+1} \newline(D) 9t+9t+3 \frac{9 t+9}{t+3}
  1. Factor out common terms: Simplify the numerator and the denominator by factoring out common terms.\newlineThe numerator 81t2+27t81t^2 + 27t can be factored by taking out the common factor of 27t27t, which gives us 27t(3t+1)27t(3t + 1).\newlineThe denominator 9t2+9t9t^2 + 9t can be factored by taking out the common factor of 9t9t, which gives us 9t(t+1)9t(t + 1).\newlineSo, the expression becomes 27t(3t+1)9t(t+1)\frac{27t(3t + 1)}{9t(t + 1)}.
  2. Cancel out common factors: Simplify the expression by canceling out common factors.\newlineWe can cancel out the common factor of 9t9t from the numerator and the denominator.\newlineThis leaves us with (3(3t+1)t+1)\left(\frac{3(3t + 1)}{t + 1}\right).
  3. Expand remaining expression: Expand the remaining expression in the numerator.\newlineMultiplying 33 by each term in the parentheses gives us 9t+39t + 3.\newlineSo, the expression is now (9t+3)/(t+1)(9t + 3)/(t + 1).
  4. Match with answer choices: Match the simplified expression with the given answer choices.\newlineThe expression (9t+3)/(t+1)(9t + 3)/(t + 1) matches with choice (C).

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