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Gwar [14]
2 years ago
8

Robert takes a roast out of the oven when the internal temperature of the roast is 165°F. After 15 minutes, the temperature of t

he roast drops to 135°F.
The temperature of the room is 70°F.

How long does it take for the temperature of the roast to drop to 110°F?
Mathematics
1 answer:
faust18 [17]2 years ago
6 0

Step-by-step explanation:

Rate of drop of temperature = Change in temperature/Rate

=> (165 - 135)/15

=> 30/15

=> 2 ⁰F/min

Now, The time at which the temperature of will be 70⁰F = 70/Rate

=> 70/2

=> 35 min

Time for 110⁰ F

=> 110/2

=> 55 min

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Answer:

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Step-by-step explanation:

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Pls need help 100 points and crown if right!!!
nekit [7.7K]

Answer:

\sf  2\dfrac{1}{4}\ cups \ of \ sugar

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3/4 cup of sugar required to make 1/3 batch of cookies

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const2013 [10]

Answer:

\rm\displaystyle  0,\pm\pi

Step-by-step explanation:

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===========================

we want to find all possible values of α+β+γ when <u>tanα+tanβ+tanγ = tanαtanβtanγ</u><u> </u>to do so we can use algebra and trigonometric skills first

cancel tanγ from both sides which yields:

\rm\displaystyle  \tan( \alpha )  +  \tan( \beta ) =  \tan( \alpha )  \tan( \beta )  \tan( \gamma )  -  \tan( \gamma )

factor out tanγ:

\rm\displaystyle  \tan( \alpha )  +  \tan( \beta ) =   \tan( \gamma ) (\tan( \alpha )  \tan( \beta ) -  1)

divide both sides by tanαtanβ-1 and that yields:

\rm\displaystyle   \tan( \gamma ) =  \frac{ \tan( \alpha )  +  \tan( \beta ) }{ \tan( \alpha )  \tan( \beta )    - 1}

multiply both numerator and denominator by-1 which yields:

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\rm\displaystyle   \tan( \gamma ) =   -  \tan( \alpha  +  \beta )

let α+β be t and transform:

\rm\displaystyle   \tan( \gamma ) =   -  \tan( t)

remember that tan(t)=tan(t±kπ) so

\rm\displaystyle   \tan( \gamma ) =    -\tan(   \alpha   +\beta\pm k\pi )

therefore <u>when</u><u> </u><u>k </u><u>is </u><u>1</u> we obtain:

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recall that if we have common trigonometric function in both sides then the angle must equal which yields:

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ratelena [41]
328 is the answer sir
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