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Alex17521 [72]
3 years ago
15

Hannah is 2 times her sister’s age. The sum of their ages is no more than 18 years. 1) Write an inequality that can be used to r

epresent this situation. 2) Using the inequality you just found, solve it to find the oldest age Hannah's sister can be.
Mathematics
1 answer:
svetoff [14.1K]3 years ago
3 0

Answer:

1) x+2x\leq 18.

2) 6 years

Step-by-step explanation:

Let Hanna's sister be x years old.


Then Hanna will be 2x years old.


The sum of their ages will be x+2x years.

The statement "no more than 18" means the sum of their ages is less than or equal to 18.

The required inequality is;

x+2x\leq 18.

2)

The inequality we just found is


x+2x\leq18.

We simplify to get;

3x\leq 18.

we divide both sides by 3 to get;

x\leq 6.

Therefore the oldest age Hannah's sister can be is 6 years




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

T(t) = 75 + 110e^{(1/30)\ln(15/22)t} \\
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(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}

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