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3 years ago

PLEASE HELP!!! Name the postulate or theorem that you can use to prove.....

Mathematics

1 answer:

bearhunter [10]3 years ago

5 0

Answer:

HL theorem

Step-by-step explanation:

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A. 29<br> B. 24<br> C. 20<br> D. 14

slava [35]

The answer is C. Hope that helps

3 0

3 years ago

Read 2 more answers

On a cold winter moming in Colorado, the outdoor temperature was 5 degrees Fahrenheit at sunrise For the next sex hours the temp

Temka [501]

Answer: 6

Step-by-step explanation:

because i said so

8 0

3 years ago

Solve 5x-3/2 + x+7/3=15

inessss [21]

The answer is x = 85/36.

7 0

3 years ago

PLEASE HELP ME !I NEED IT

Shalnov [3]

Answer:

Step-by-step explanation:

3 0

3 years ago

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.

4 0

3 years ago