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T(t)=25+65*(0.78)^(t) A pot of soup is heated and then left to cool in a room with a constant temperature. The equation gives the temperature of the soup, T(t), in degrees Celsius (^(@)C),t minutes after it is heated. What is the initial temperature of the soup before it begins to cool? Choose 1 answer: (A) 25^(@)C (B) 65^(@)C (c) 78^(@)C (D) 90^(@)C

T(t)=25+65⋅(0.78)t T(t)=25+65 \cdot(0.78)^{t} \newlineA pot of soup is heated and then left to cool in a room with a constant temperature. The equation gives the temperature of the soup, T(t) T(t) , in degrees Celsius (∘C),t \left({ }^{\circ} \mathrm{C}\right), t minutes after it is heated. What is the initial temperature of the soup before it begins to cool?\newlineChoose 11 answer:\newline(A) 25∘C 25^{\circ} \mathrm{C} \newline(B) 65∘C 65^{\circ} \mathrm{C} \newline(C) 78∘C 78^{\circ} \mathrm{C} \newline(D) 90∘C 90^{\circ} \mathrm{C}

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Q. T(t)=25+65⋅(0.78)t T(t)=25+65 \cdot(0.78)^{t} \newlineA pot of soup is heated and then left to cool in a room with a constant temperature. The equation gives the temperature of the soup, T(t) T(t) , in degrees Celsius (∘C),t \left({ }^{\circ} \mathrm{C}\right), t minutes after it is heated. What is the initial temperature of the soup before it begins to cool?\newlineChoose 11 answer:\newline(A) 25∘C 25^{\circ} \mathrm{C} \newline(B) 65∘C 65^{\circ} \mathrm{C} \newline(C) 78∘C 78^{\circ} \mathrm{C} \newline(D) 90∘C 90^{\circ} \mathrm{C}
  1. Determine Initial Temperature: To find the initial temperature of the soup, we need to determine the value of T(t)T(t) at t=0t = 0, which is when the soup starts cooling.
  2. Substitute t=0t = 0: Substitute t=0t = 0 into the equation T(t)T(t) to find the initial temperature.\newlineT(0)=25+65×(0.78)0T(0) = 25 + 65 \times (0.78)^0
  3. Calculate (0.78)0(0.78)^0: Calculate the value of (0.78)0(0.78)^0. Any non-zero number raised to the power of 00 is 11.(0.78)0=1(0.78)^0 = 1
  4. Substitute (0.78)0(0.78)^0: Now, substitute the value of (0.78)0(0.78)^0 back into the equation to find T(0)T(0).\newlineT(0)=25+65×1T(0) = 25 + 65 \times 1
  5. Perform Multiplication and Addition: Perform the multiplication and addition to find the initial temperature.\newlineT(0)=25+65T(0) = 25 + 65\newlineT(0)=90T(0) = 90

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