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STALIN [3.7K]
3 years ago
13

(-3/4)^5 Please help ASAP!!!

Mathematics
2 answers:
zhuklara [117]3 years ago
4 0
-243/1024 -3^5/4^5 = -243/4^5 which equals to -243/1024
coldgirl [10]3 years ago
3 0

Answer:

-243/1024

Step-by-step explanation:

(-3/4)x(-3/4)x(-3/4)x(-3/4)x(-3/4) = -243/1024

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At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel
Alexeev081 [22]
Use Law of Cooling:
T(t) = (T_0 - T_A)e^{-kt} +T_A
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
42 = (80-20)e^{-3k} +20 \\  \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\  \\ k = -\frac{1}{3} ln (\frac{11}{30})

Substituting k back into cooling equation gives:
T(t) = 60(\frac{11}{30})^{t/3} + 20

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\  \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80
Solve for x:
x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80
Sub back into original cooling equation, x = T(t)
-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20
Solve for t:
60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3}  = 60 \\  \\  (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\  \\  \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\  \\ \\  t = 4.959

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM
8 0
3 years ago
Give answer for this question ❓​
rjkz [21]

Answer:

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Step-by-step explanation:

B is false because rational numbers have either repeating decimals or numbers that come to an end.

7 0
3 years ago
Will give brainliest and add more points to your acc if good awnser Please help ASAP
joja [24]
Hailey's mixing two different coffee blends.  Represent them by x and y (in pounds).  Then x + y = 5 lb, and x = 5 - y.How much puree Sum. beans are we talking about here?
0.20x +0.80y = 0.60(5 lb)  Mult all 3 terms by 100 to get rid of factions:
20x + 80 y = 300.  Substitute 5-y for x:
20(5-y) + 80y = 300 => 100-20y + 80y = 300 => 60y = 200,  so                                                                                  y = 20/6 or 10/3 lb                                              den x = 5-10/3, or    x =5/3 lb
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