Submitted by gemini on Sun, 05/10/2020 - 03:02 pm.

**Login**

No account yet? » Register

Can't login? » Reset your password

## New forum topics

- At 2:00 pm,thermometer reading 80oF.itis taken outside where the air temp is 20oF. At 2:03 pm, the temp reading yielded by the thermometer is 42oF. Later it was brought inside at 80oF. At 2:10 the reading is 71oF. When was the thermometer brought indoors?
- What 'positive' values of x, y, and z that will make x/(y+z) + y/(z+x) + z/(x+y) = 4?
- A man wishes his son to receive P200,000 ten years from now. What amount should he invest now if it will earn interest of 10% compounded annually during the first 5 years and 12% compounded quarterly during the next 5 years.
- Among 10000 random digits, find the probability that the 3 appears at most 950 times.
- Is this a rule in semi circles?

## Active forum topics

- At 2:00 pm,thermometer reading 80oF.itis taken outside where the air temp is 20oF. At 2:03 pm, the temp reading yielded by the thermometer is 42oF. Later it was brought inside at 80oF. At 2:10 the reading is 71oF. When was the thermometer brought indoors?
- Find the differential equation of family of circles with center on the line y= -x and passing through the origin.
- What 'positive' values of x, y, and z that will make x/(y+z) + y/(z+x) + z/(x+y) = 4?
- A man wishes his son to receive P200,000 ten years from now. What amount should he invest now if it will earn interest of 10% compounded annually during the first 5 years and 12% compounded quarterly during the next 5 years.
- Among 10000 random digits, find the probability that the 3 appears at most 950 times.

## $T = T_s + (T_o - T_s)e^{-kt…

$T = T_s + (T_o - T_s)e^{-kt}$

At 2:00PM

$T = 20 + (80 - 20)e^{-kt}$

$T = 20 + 60e^{-kt}$

At 2:03PM

$42 = 20 + 60e^{-k(3)}$

$e^{-3k} = \frac{11}{30}$

$e^{-k} = \left( \frac{11}{30} \right)^{1/3}$

Hence,

$T = 20 + 60\left( \frac{11}{30} \right)^{t/3}$

t

_{1}minutes after 2:03, the thermometer reads$T = 20 + 60\left( \frac{11}{30} \right)^{{t_1}/3}$

When the thermometer is brought back to the room

$T = 80 + (T_o - 80)\left( \frac{11}{30} \right)^{t/3}$

$T = 80 + \left[ 20 + 60\left( \frac{11}{30} \right)^{{t_1}/3} - 80 \right] \left( \frac{11}{30} \right)^{t/3}$

$T = 80 + \left[ 60\left( \frac{11}{30} \right)^{{t_1}/3} - 60 \right] \left( \frac{11}{30} \right)^{t/3}$

$T = 80 + 60 \left[ \left( \frac{11}{30} \right)^{{t_1}/3} - 1 \right] \left( \frac{11}{30} \right)^{t/3}$

At 2:10PM, t = 7 - t

_{1}$71 = 80 + 60 \left[ \left( \frac{11}{30} \right)^{{t_1}/3} - 1 \right] \left( \frac{11}{30} \right)^{(7 - t_1)/3}$

$-9 = 60 \left[ \left( \frac{11}{30} \right)^{t_1/3} - 1 \right] \dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{t_1/3}}$

$-9 = 60 \left[ \left( \frac{11}{30} \right)^{7/3} - \dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{t_1/3}} \right]$

$\dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{t_1/3}} = \left( \frac{11}{30} \right)^{7/3} + \dfrac{9}{60}$

$\dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{7/3} + \dfrac{9}{60}} = \left( \frac{11}{30} \right)^{t_1/3}$

$\left[ \dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{7/3} + \dfrac{9}{60}} \right]^3 = \left( \frac{11}{30} \right)^{t_1}$

$\ln \left[ \dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{7/3} + \dfrac{9}{60}} \right]^3 = \ln \left( \frac{11}{30} \right)^{t_1}$

$\ln \left[ \dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{7/3} + \dfrac{9}{60}} \right]^3 = t_1 \ln \left( \frac{11}{30} \right)$

$t_1 = 2.8 ~ \text{min}$

time = 2:05:48.6PM