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Ahat [919]
3 years ago
5

Gina bought 2.3 pounds of red apples and 2.42 pounds of green apples.there were on sale for 0.75 a pound.how much did the apples

cost altogether
Mathematics
2 answers:
shutvik [7]3 years ago
3 0
2.3 + 2.42 =  4.72lbs     
4.72 x .75 =  $3.54
padilas [110]3 years ago
3 0

Answer:  The required cost of the apples altogether is $3.54.

Step-by-step explanation:  Given that Gina bought 2.3 pounds of red apples and 2.42 pounds of green apples.there were on sale for 0.75 a pound.

We are to find the cost of the apples altogether.

Total quantity of apples bought by Gina is

Q = (2.3 + 2.42) pounds = 4.72 pounds.

The cost of 1 pound of apple = $ 0.75.

Therefore, the cost of 4.72 pounds of apple is given by

$ (4.72 × 0.75) = $ 3.54.

Thus, the required cost of the apples altogether is $3.54.

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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}

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