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

Find Initial Temperature: To find the initial temperature of the soup, we need to determine the value of T(t) at t = 0, which is when the soup was just heated and before it begins to cool.Substitute t = 0: Substitute t = 0 into the equation T(t) to find the initial temperature. T(0) = 25 + 65 * (0.78)^0Calculate (0.78)^0: Calculate the value of (0.78)^0. Any non-zero number raised to the power of 0 is 1. (0.78)^0 = 1Substitute value back: Now, substitute the value back into the equation to find T(0). T(0) = 25 + 65 * 1Perform multiplication and addition: Perform the multiplication and addition to find the initial temperature. T(0) = 25 + 65 T(0) = 90

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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

$T(t)=25+65 \cdot(0.78)^{t}$ $\newline$ 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 $\left({ }^{\circ} \mathrm{C}\right), t$ minutes after it is heated. What is the initial temperature of the soup before it begins to cool? $\newline$ Choose $1$ answer: $\newline$ (A) $25^{\circ} \mathrm{C}$ $\newline$ (B) $65^{\circ} \mathrm{C}$ $\newline$ (C) $78^{\circ} \mathrm{C}$ $\newline$ (D) $90^{\circ} \mathrm{C}$

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Algebra 1

Transformations of quadratic functions

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T(t)=25+65 \cdot(0.78)^{t}

\newline

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

\left({ }^{\circ} \mathrm{C}\right), t

minutes after it is heated. What is the initial temperature of the soup before it begins to cool?

\newline

Choose

1

answer:

\newline

(A)

25^{\circ} \mathrm{C}

\newline

(B)

65^{\circ} \mathrm{C}

\newline

(C)

78^{\circ} \mathrm{C}

\newline

(D)

90^{\circ} \mathrm{C}

Find Initial Temperature: To find the initial temperature of the soup, we need to determine the value of $T(t)$ at $t = 0$ , which is when the soup was just heated and before it begins to cool.
Substitute $t = 0$ : Substitute $t = 0$ into the equation $T(t)$ to find the initial temperature. $\newline$ $T(0) = 25 + 65 \times (0.78)^0$
Calculate $(0.78)^0$ : Calculate the value of $(0.78)^0$ . Any non-zero number raised to the power of $0$ is $1$ . $(0.78)^0 = 1$
Substitute value back: Now, substitute the value back into the equation to find $T(0)$ . $\newline$ $T(0) = 25 + 65 \times 1$
Perform multiplication and addition: Perform the multiplication and addition to find the initial temperature. $\newline$ $T(0) = 25 + 65$ $\newline$ $T(0) = 90$