Simplify: x2 + 5x + 6(x + 3) x 42x + 4 4 2 2(x + 2) (x + 2)

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

7 0

3 years ago

Your next math test is worth 111 points and contains 32 problems. Each problem is worth either 4 points or 3 points. How many 3

Dafna11 [192]

We call x the 4 points-worth problems and y the 3 points-worth problems

You know that x+y = 32

4x+3y = 111

You know that the difference from the x and y is 32 so write:

x = 32-y

Substitute at x the value of 32-y

4(32-y)+3y = 111

128-4y +3y = 111

-4y+3y = 111-128

-y = -17

y = 17

3 0

3 years ago

Read 2 more answers

The random variable X is exponentially distributed, where X represents the waiting time to be seated at a restaurant during the

erastova [34]

Answer:

The probability that the wait time is greater than 14 minutes is 0.4786.

Step-by-step explanation:

The random variable X is defined as the waiting time to be seated at a restaurant during the evening.

The average waiting time is, β = 19 minutes.

The random variable X follows an Exponential distribution with parameter $\lambda=\frac{1}{\beta}=\frac{1}{19}$ .

The probability distribution function of X is:

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

Compute the value of the event (X > 14) as follows:

$P(X>14)=\int\limits^{\infty}_{14} {\lambda e^{-\lambda x}} \, dx=\lambda \int\limits^{\infty}_{14} {e^{-\lambda x}} \, dx\\=\lambda |\frac{e^{-\lambda x}}{-\lambda}|^{\infty}_{14}=e^{-\frac{1}{19} \times14}-0\\=0.4786$

Thus, the probability that the wait time is greater than 14 minutes is 0.4786.

7 0

2 years ago

In a company, 80% of the workers are women. if 880 men work for the company, how many workers are there in all?

ELEN [110]

If 80% are women, then 20% are men

20% of what equals 880
0.20x = 880
x = 880/.20
x = 4400

check..
men : 20% of 4400 = 0.20(400) = 880
women : 80% of 4400 = 0.80(4400) = 3520
totaling : 3520 + 880 = 4400

so there are a total of 4400 workers

6 0

3 years ago

PLEASE HELP! I AM CRYING AND NEED HELP :'(

Svetllana [295]

Answer:

9300

Step-by-step explanation:

To find the total surface are of 150 boxes, all we have to do is multiply the surface area of one box ( 62 which was calculated on this question brainly.com/question/23446865 ) by the total amount of boxes (150)

So your answer = 150 * 62 which equals 9300

4 0

3 years ago