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3 years ago

Cos^2x-sin^2x/sin^2x+sinxcosx=cotx-1

Mathematics

1 answer:

meriva3 years ago

8 0

Answer:

$\bold{\frac{(cosx-sinx)}{(sinx)}}=\bold{\frac{cosx-sinx}{sinx}}$

Step-by-step explanation:

$\frac{cos^2x-sin^2x}{sin^2x+sinxcosx}=cotx-1$

We're going to start by manipulating the left side of the equation and making it the same form as $cotx-1$ .

Start by applying the difference of two squares formula to the numerator, like so:

$\frac{(cosx+sinx)(cosx-sinx)}{sin^2x+sinxcosx}$

Now simplify the denominator by expanding the $sin^2x$ .

$\frac{(cosx+sinx)(cosx-sinx)}{(sinx)(sinx)+sinxcosx}$

The denominator can even be further simplified since both addends (when added together = a sum) have the common factor of $sinx$ . Factor it out.

$\frac{(cosx+sinx)(cosx-sinx)}{(sinx)(sinx+cosx)}$

Cancel out the common factor $(cosx+sinx)$ .

$\bold{\frac{(cosx-sinx)}{(sinx)}}$

Since this is the furthest simplified that the left side can be manipulated, let's see if can try to manipulate the right side to also look like $\frac{(cosx-sinx)}{(sinx)}$ .

Start by expressing $cotx-1$ with $sinx$ and $cosx$ , since we know that cotangent is simply $\frac{x}{y} \rightarrow\frac{cosx}{sinx}$ .

$\frac{cosx}{sinx}-1$

We can simplify this expression to look like our expression we found by manipulating the left side $(\frac{(cosx-sinx)}{(sinx)})$ by making the 1 have a common denominator of $sinx$ .

To do this, multiply 1 by $\frac{sinx}{sinx}$ . Now the expression should look like:

$\frac{cosx}{sinx}-\frac{sinx}{sinx}$

Since they have a common denominator we can write the expression under one fraction, like so:

$\bold{\frac{cosx-sinx}{sinx}}$

This looks exactly the same as what we manipulated the left side to be $(\frac{(cosx-sinx)}{(sinx)})$ , just without parentheses. I put both expressions in bold. Therefore, this identity proves to be true as we just proved it.

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We know that the area of the smaller figure is $25in^{2}$ , and its volume is $250in^{3}$ . We also know that the area of the larger figure is $36in^{2}$ ; since we don't now its volume, lets represent it with $X$ :
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3 years ago

Read 2 more answers

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3 years ago

Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie

ludmilkaskok [199]

Answer:

$\lambda \geq 6.63835$

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that $X \sim Poisson(\lambda)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda$

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

$P(X\geq 2)=1-P(X$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}$

$P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}$

And replacing we have this:

$P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]$

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)$

And we want this probability that at least of 99%, so we can set upt the following inequality:

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99$

And now we can solve for $\lambda$

$0.01 \geq e^{-\lambda}(1+\lambda)$

Applying natural log on both sides we have:

$ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)$

$ln(0.01) \geq -\lambda+ln(1+\lambda)$

$\lambda-ln(1+\lambda)+ln(0.01) \geq 0$

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

$x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}$

Where :

$f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)$

$f'(x_n)=1-\frac{1}{1+\lambda}$

Iterating as shown on the figure attached we find a final solution given by:

$\lambda \geq 6.63835$

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3 years ago

Helllppppp can’t find the answer!!!

alukav5142 [94]

Slope of a line can be determined using this formula
m = (y₂ - y₁) / (x₂ - x₁)

From the question, we know that
(x₁,y₁) = (2,-3)
(x₂,y₂) = (2,9)

plug the numbers into the formula
m = (y₂ - y₁) / (x₂ - x₁)
m = (9 - (-3)) / (2 - 2)
m = (9 + 3) / (2 - 2)
m = 12/0
m = undefined
The pairs must form a vertical line

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3 years ago