I need this answered as soon as possible

Mathematics

2 answers:

Andre45 [30]3 years ago

8 0

Answer:

the condition of having an abnormally (typically dangerously) low body temperature

Step-by-step explanation:

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sdas [7]3 years ago

3 0

Step-by-step explanation:

Let

Cost of popcorn = x

Cost of drink = y

3x + 5y = 24.5(1)

4x + 3y = 23.5(2)

From 1

3x + 5y = 24.5

3x = 24.5 - 5y

x = (24.5 - 5y)/3

Putting value of x in 2

4(24.5 - 5y)/3 + 3y = 23.5

98 - 20y/3 + 3y = 23.5

98 - 20y + 9y/3 = 23.5

98 - 11y = 3 × 23.5

98 - 11y = 70.5

-11y = 70.5 - 98.5

-11y = -28

y = 2.5

3x + 5(2.5) = 24.5

3x + 12.5 = 24.5

3x = 24.5 - 12.5

3x = 12

x = 12/3

x = 4

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How do you this question (pre-cal)

Novay_Z [31]

The goal to proving identities is to transform one side into the other. We can only pick one side to transform while the other side stays the same the entire time. The general rule of thumb is to transform the more complicated side (though there may be exceptions to this guideline).

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:

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