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3 years ago

Factor 2x4 - 20x2 - 78.

Mathematics

2 answers:

Komok [63]3 years ago

8 0

Answer:

2(x² + 3)(x² - 13)

Step-by-step explanation:

2x⁴ - 20x² - 78

Factor out 2.

2(x⁴ - 10x² - 39)

Find 2 numbers that multiply to get -39 and add to get -10. Those numbers are -13 and 3.

2(x⁴ - 13x² + 3x² - 39)

2(x² + 3)(x² - 13)

Airida [17]3 years ago

7 0

Answer:

2x⁴ - 20x² - 78

To factor the expression look for the LCM of the numbers

LCM of the numbers is 2

Factorize that one out

That's

2( x⁴ - 10x² - 39)

Hope this helps

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Answer:

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3 years ago

What is the answer to 3x+x/4=x+3/4

erica [24]

Answer:x = 1

Step-by-step explanation:

3x + x/4 = x + 3/4

Subtract x from each side

2x + x/4 = 3/4

Times each side by 4

The 4 beneath the x cancels out

3/4 x 4 = 3

2x + x = 3

3x = 3

Divide each side by 3

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5 0

3 years ago

If you drive 23 miles south, make a turn and drive 39 miles east, how far are you, in a straight line, from

Rom4ik [11]

Answer:

45.27 miles

Step-by-step explanation:

Pythagorean theorem

23^2 + 39^2 = d ^ 2

d = 45.27 miles

6 0

2 years ago

Can anyone help me with this problem? If you can include the steps, please do! Thanks :)

8090 [49]

Answer:1=70 2=65 3=95

Step-by-step explanation:

Angles in a triangle add up to 180°

180-45=135-65=70

1=70°

Angles on a straight line add up to 180°

180-45=135-70=65

2=65°

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3 0

3 years ago

The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β =

stich3 [128]

I'm assuming $\alpha$ is the shape parameter and $\beta$ is the scale parameter. Then the PDF is

$f_X(x)=\begin{cases}\dfrac29xe^{-x^2/9}&\text{for }x\ge0\\\\0&\text{otherwise}\end{cases}$

a. The expectation is

$E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\frac29\int_0^\infty x^2e^{-x^2/9}\,\mathrm dx$

To compute this integral, recall the definition of the Gamma function,

$\Gamma(x)=\displaystyle\int_0^\infty t^{x-1}e^{-t}\,\mathrm dt$

For this particular integral, first integrate by parts, taking

$u=x\implies\mathrm du=\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X]=\displaystyle-xe^{-x^2/9}\bigg|_0^\infty+\int_0^\infty e^{-x^2/9}\,\mathrm x$

$E[X]=\displaystyle\int_0^\infty e^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ , so that $\mathrm dx=\dfrac32y^{-1/2}\,\mathrm dy$ :

$E[X]=\displaystyle\frac32\int_0^\infty y^{-1/2}e^{-y}\,\mathrm dy$

$\boxed{E[X]=\dfrac32\Gamma\left(\dfrac12\right)=\dfrac{3\sqrt\pi}2\approx2.659}$

The variance is

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2$

The second moment is

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f_X(x)\,\mathrm dx=\frac29\int_0^\infty x^3e^{-x^2/9}\,\mathrm dx$

Integrate by parts, taking

$u=x^2\implies\mathrm du=2x\,\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X^2]=\displaystyle-x^2e^{-x^2/9}\bigg|_0^\infty+2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

$E[X^2]=\displaystyle2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ again to get

$E[X^2]=\displaystyle9\int_0^\infty e^{-y}\,\mathrm dy=9$

Then the variance is

$\mathrm{Var}[X]=9-E[X]^2$

$\boxed{\mathrm{Var}[X]=9-\dfrac94\pi\approx1.931}$

b. The probability that $X\le3$ is

$P(X\le 3)=\displaystyle\int_{-\infty}^3f_X(x)\,\mathrm dx=\frac29\int_0^3xe^{-x^2/9}\,\mathrm dx$

which can be handled with the same substitution used in part (a). We get

$\boxed{P(X\le 3)=\dfrac{e-1}e\approx0.632}$

c. Same procedure as in (b). We have

$P(1\le X\le3)=P(X\le3)-P(X\le1)$

and

$P(X\le1)=\displaystyle\int_{-\infty}^1f_X(x)\,\mathrm dx=\frac29\int_0^1xe^{-x^2/9}\,\mathrm dx=\frac{e^{1/9}-1}{e^{1/9}}$

Then

$\boxed{P(1\le X\le3)=\dfrac{e^{8/9}-1}e\approx0.527}$

7 0

3 years ago