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3 years ago

Greatest common factor 16, and 24

Mathematics

2 answers:

kupik [55]3 years ago

7 0

$\large\text{Hey there!}$

$\large\text{Greatest common factor (GCF) of 16 \& 24}$

$\large\text{16: 1, 2 , 4 , 8 , \& 16}$

$\large\text{24: 1 , 2 , 3 , 4 , 6 , 8 , 12 , \& 24}$

$\large\text{Like terms of both of your numbers: 1, 2, 4, \& 8}$

$\large\text{The GREATEST out of like terms is: 8}$

$\boxed{\boxed{\large\text{Thus, your answer is :\boxed{\large\text{\bf{Your Greatest Common Factor is: \underline{8}}}}}}}\checkmark$

$\text{Good luck on your assignment and enjoy your day!}$

~ $\dfrac{\frak{LoveYourselfFirst}}{:)}$

statuscvo [17]3 years ago

6 0

Answer:

The answer is 8. :D Hope this helps!

Explanation Edit: The greatest common factor is 8. Out of all the like terms, 1, 2, 4, and 8, 8 is the greatest. The greatest term is 8 = The greatest common factor. 8 is the greatest common factor.

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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

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Evgesh-ka [11]

Answer:

Hey there!

3/4= 0.75

1/5=0.2

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Thus, 1/7 is the smallest fraction.

Hope this helps :)

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