QUESTION 1 Simplify the following completely: \( 1.1 \frac{10^{2 x+3} \cdot 4^{1-x}}{25^{2+x}} \) \( 1.2 \frac{3^{t}-3^{t-2}}{2.3^{t}-3^{t}} \)

To simplify the expression \( 1.1 \frac{10^{2 x+3} \cdot 4^{1-x}}{25^{2+x}} \), first, rewrite the bases in terms of powers of 10 and 5. Notice that \( 10 = 10^1 \), \( 4 = 2^2 \), and \( 25 = 5^2 \). This allows us to transform the expression. Next, substitute: \[ 25^{2+x} = (5^2)^{2+x} = 5^{4 + 2x} \] Continuing the simplification gives: \[ \frac{10^{2x+3} \cdot 4^{1-x}}{25^{2+x}} = \frac{10^{2x+3} \cdot (2^2)^{1-x}}{5^{4 + 2x}} = \frac{10^{2x+3} \cdot 2^{2-2x}}{5^{4 + 2x}} \] Using \(10 = 2 \cdot 5\), we can express \(10^{2x+3}\) as: \[ (2 \cdot 5)^{2x+3} = 2^{2x+3} \cdot 5^{2x+3} \] Plugging this back in gives: \[ \frac{2^{2x+3} \cdot 5^{2x+3} \cdot 2^{2-2x}}{5^{4+2x}} = \frac{2^{2x+3+2-2x} \cdot 5^{2x+3}}{5^{4+2x}} = \frac{2^5 \cdot 5^{2x+3}}{5^{4 + 2x}} = 32 \cdot 5^{(2x+3) - (4 + 2x)} = 32 \cdot 5^{-1} = \frac{32}{5} \] For \( 1.2 \frac{3^{t}-3^{t-2}}{2.3^{t}-3^{t}} \), observe that you can factor the numerator and the denominator. Starting with the numerator: \[ 3^t - 3^{t-2} = 3^{t-2}(3^2 - 1) = 3^{t-2} \cdot 8 \] And proceeding with the denominator: \[ 2 \cdot 3^{t} - 3^{t} = (2 - 1) \cdot 3^{t} = -3^{t} \] Rearranging the expression: \[ \frac{3^{t-2} \cdot 8}{-3^{t}} = -\frac{8 \cdot 3^{t-2}}{3^t} = -\frac{8}{3^2} = -\frac{8}{9} \] So the answers are: 1.1: \(\frac{32}{5}\) 1.2: \(-\frac{8}{9}\)