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Ivahew [28]
4 years ago
5

Find the distance between the points with the given coordinates (-2, -5) and (-2, 4).

Mathematics
1 answer:
bija089 [108]4 years ago
6 0
I hope this helps you

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A roast turkey is taken from an oven when its temperature has reached 185 degrees F and is placed on a table in a room where the
Vedmedyk [2.9K]
(a) Using Newton's Law of Cooling, \dfrac{dT}{dt} = k(T - T_s), we have \dfrac{dT}{dt} = k(T - 75) where T is temperature after T minutes.
Separate by dividing both sides by T - 75 to get \dfrac{dT}{T - 75} = k dt. Integrate both sides to get \ln|T - 75| = kt + C.

Since T(0) = 185, we solve for C:
|185 - 75| = k(0) + C\ \Rightarrow\ C = \ln 110
So we get \ln|T - 75| = kt + \ln 110. Use T(30) = 150 to solve for k:
\ln| 150 - 75 | = 30k + \ln 110\ \Rightarrow\ \ln 75 - \ln 110= 30k \Rightarrow \\ k= \frac{1}{30}\ln (75/110) = \frac{1}{30}\ln(15/22)

So

\ln|T - 75| = kt + \ln 110 \Rightarrow |T - 75| = e^{kt + \ln110} \Rightarrow \\ \\
|T - 75| = 110e^{kt} \Rightarrow T - 75 = \pm110e^{(1/30)\ln(15/22)t}  \Rightarrow \\
T = 75 \pm110e^{(1/30)\ln(15/22)t}

But choose Positive because T > 75. Temp of turkey can't go under.

T(t) = 75 + 110e^{(1/30)\ln(15/22)t} \\
T(45) = 75 + 110e^{(1/30)\ln(15/22)(45)}  = 136.929 \approx 137{}^{\circ}F

(b)

T(t) = 75 + 110e^{(1/30)\ln(15/22)t} = 75 + 110(15/22)^{t/30}  \\
100 = 75 + 110(15/22)^{t/30}   \\
25 = 110(15/22)^{t/30}  
\frac{25}{110} = (15/22)^{t/30}   \\
\ln(25/110) / ln(15/22) = t/30 \\
t = 30\ln(25/110) / ln(15/22)  \approx 116\ \mathrm{min}

Dogs of the AMS.
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3 years ago
It takes Jimmy 11 minutes to sweep a porch.
lara31 [8.8K]
Like 6 minutes or somthin
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What is the value of x in the equation x - ty = 30, when y = 15?
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7 0
3 years ago
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Explain why there must be at least two lines on any given plane.
Ghella [55]
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6 0
3 years ago
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The temperature at 6.00 am was - 12 the temperature increased by 1/2 each hour for 6 hours what was the temperature in degrees F
Illusion [34]

Answer: -9 degrees Fahrenheit

Step-by-step explanation:

Given: Temperature at 6:00 AM = -12 degrees Fahrenheit

Temperature increased each hour = \dfrac12 degrees Fahrenheit

Temperature increase in 6 hours = 6\times\dfrac12=3\text{ degrees Fahrenheit}

Temperature at noon = Temperature at 6:00 AM+Temperature increase in 6 hours

= -12+3 degrees Fahrenheit

= -9 degrees Fahrenheit

Hence, the temperature in degrees Fahrenheit at noon= -9 degrees Fahrenheit

8 0
3 years ago
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