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3 years ago

What is the measure of angle LKJ? Round to the nearest whole degree. 41° 45° 49° 55°

Mathematics

2 answers:

Sonbull [250]3 years ago

5 0

Answer:49

Step-by-step explanation:

Over [174]3 years ago

5 0

Answer:

Step-by-step explanation:

jus took test on EDGE! :)

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| 4x +2|=10 confirm your solution using a graph or table

klio [65]

Remove the absolute value term. 4x+2=10 solve this, x=2

Set up the negative portion 4x+2= -10 solve this, x= -3

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sukhopar [10]

Answer:

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Step-by-step explanation:

<em>Excuse me if i'm wrong, it seems like you have the same test or almost. And this was the answer on mine. I hope it works out for you!</em>

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3 years ago

Does water have a taste

MariettaO [177]

Answer:

nope

Step-by-step explanation:

Water is tasteless and odorless

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3 years ago

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A bottle of hand lotion is on sale for 2.25.if the price represents a 50percent discount from the original price,what is the ori

Flura [38]

Answer: $4.50

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3 years ago

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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3 years ago