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Rainbow [258]
3 years ago
12

your turkey must cook for 5 & a half hours. You want to eat at 4 o'clock.It will take 15 minutes to carve the turkey. At wha

t time will the turkey going to the oven?!
Mathematics
2 answers:
kkurt [141]3 years ago
5 0
The turkey will go into the oven at about 10:15
hope ti helps
and choose brainliest answer
Lubov Fominskaja [6]3 years ago
3 0

Answer:

The turkey should go to the oven at 10:15

Step-by-step explanation:

1. As you want to eat at 4 o´clock you should calculate how many time you will spend cooking and carving the turkey, so:

Time cooking the turkey = 5.5 hours

Time carving the turkey = 15minutes*\frac{1hour}{60minutes}=0.25hours

Total time preparing the turkey = 5.5hours + 0.25hours

Total time preparing the turkey = 5.75 hours

2. Subtract the total time that you will spend preparing the turkey from the 16 hours:

16 - 5.75 = 10.25hours

0.25hours*\frac{60minutes}{1hour}=15minutes

Therefore, the turkey should go to the oven at 10:15

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So I'll take the left hand side and try to turn it into \csc^2( B )

One way we can do that is through the following steps:

\frac{\tan(B) + \cot(B)}{\tan(B)} = \csc^2(B)\\\\\frac{\tan(B)}{\tan(B)} + \frac{\cot(B)}{\tan(B)} = \csc^2(B)\\\\1 + \cot(B)*\frac{1}{\tan(B)} = \csc^2(B)\\\\1 + \cot(B)*\cot(B) = \csc^2(B)\\\\1 + \cot^2(B) = \csc^2(B)\\\\1 + \frac{cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{sin^2(B)}{\sin^2(B)}+\frac{cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{sin^2(B)+cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{1}{\sin^2(B)} = \csc^2(B)\\\\\csc^2(B)=\csc^2(B) \ \ {\Large \checkmark}\\\\

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Answer:

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