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Bingel [31]
3 years ago
6

Whats the intrgral of

D%20" id="TexFormula1" title=" \int \frac{x^2+x-3}{(x^3+x^2-4x-4)^2} " alt=" \int \frac{x^2+x-3}{(x^3+x^2-4x-4)^2} " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
rosijanka [135]3 years ago
5 0
\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx

Notice that x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x-2)(x+2)(x+1). Decompose the integrand into partial fractions:

\dfrac{x^2+x-3}{(x-2)^2(x+2)^2(x+1)^2}
=\dfrac1{3(x+1)}-\dfrac{11}{32(x+2)}-\dfrac1{3(x+1)^2}-\dfrac1{16(x+2)^2}+\dfrac1{96(x-2)}+\dfrac1{48(x-2)^2}

Integrating term-by-term, you get

\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx
=-\dfrac1{48(x-2)}+\dfrac1{3(x+1)}+\dfrac1{16(x+2)}+\dfrac1{96}\ln|x-2|+\dfrac13\ln|x+1|-\dfrac{11}{32}\ln|x+2|+C
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Which equation involves a prime quadratic and cannot be solved by factoring? A. x2 + 5x − 4 = 0 B. x2 − x − 6 = 0 C. x2 + 3x − 4
Damm [24]

Answer:

Option A

Step-by-step explanation:

We can try to factorize all the options one by one.

For A:

x^{2} +5x-4\\=> x^{2}+4x+x-4\\=>x(x+4)+1(x-4)

We can see that the quadratic expression cannot be solved by factorization as the factors at the end of factorization are not equal in both brackets. So Option A is the correct answer for the given question.

Moreover we can also note that all the other quadratic expressions can be factorized ..

8 0
3 years ago
(Can someone help I'm taking a math quiz.) Seventy parents showed up for a parent meeting at school. Mrs. Quinn set up 12 tables
il63 [147K]

Answer:

for every 2 tables there are 10 chairs, so for 12 tables there are 60 chairs

4 0
2 years ago
Read 2 more answers
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

4 0
2 years ago
Drum Tight Containers is designing an​ open-top, square-based, rectangular box that will have a volume of 62.5 in cubed. What di
Orlov [11]
Let the square base of the container be of side s inches and the height of the container be h inches, then
Surface are of the container, A = s^2 + 4sh
For minimum surface area, dA / ds + dA / dh = 0
i.e. 2s + 4h + 4s = 0
6s + 4h = 0
s = -2/3 h

But, volume of container = 62.5 in cubed
i.e. s^2 x h = 62.5
(-2/3 h)^2 x h = 62.5
4/9 h^2 x h = 62.5
4/9 h^3 = 62.5
h^3 = 62.5 x 9/4 = 140.625
h = cube root of (140.625) = 5.2 inches
s = 2/3 h = 3.47

Therefore, the dimensions of the square base of the container is 3.47 inches and the height is 5.2 inches.

The minimum surface area = s^2 + 4sh = (3.47)^2 + 4(3.47)(5.2) = 12.02 + 72.11 = 84.13 square inches.

5 0
3 years ago
Read 2 more answers
Need help will give thanks and brainliest
Lynna [10]

Answer:

πr2h

Step-by-step explanation:

7 0
3 years ago
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