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Licemer1 [7]
3 years ago
10

PLEASE HELP !!!! Please help !!!

Mathematics
1 answer:
BartSMP [9]3 years ago
7 0

Answer:

1450$

Step-by-step explanation:

follow the basic steps to calculate the mean.

1. add up all of the numbers in the data set. in this case you would add up the all of the money since the question is asking for that.

2. divide by the amont of numbers there are.

1400+800+2150=4350

4350÷3=1450

hope this helped :)

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3 years ago
) f) 1 + cot²a = cosec²a​
notsponge [240]

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

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\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}

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\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}

Another identity is:

\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}

Therefore:

\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}

Another identity:

\displaystyle \large{\sin^2x+\cos^2x=1}

Therefore:

\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}\longrightarrow \boxed{ \frac{1}{\sin^2x}={\frac{1}{\sin^2x}}}

Hence proved, this is proof by using identity helping to find the specific identity.

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The answer is correct
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