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3 years ago

Find the x intercept. y = x + 3 / x - 2

Mathematics

1 answer:

VikaD [51]3 years ago

8 0

Answer:

(-3, 0).

Step-by-step explanation:

At the x intercept y = 0 so

(x + 3) / (x - 2) = 0

x + 3 = 0

x = -3.

x-intercept is at (-3,0).

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A student solved 6x+4=2(x−8) in four steps as shown:

mr_godi [17]

The first thing you should do is solve the equation yourself.

1) Distribute the 2.
6x + 4 = 2x – 16

2) Next, you'll want to get the x's on one side. So add -2x to both sides.
6x + 4 + -2x = 2x + -2x - 16
4x + 4 = -16

3) Now subtract 4 from both sides
4x + 4 – 4 = -16 – 4
4x = -12

4) Finally, divide both sides by 4
4x/4 = -12/4
x = –3

To solve this problem all you need to do is look back out you work, and figure out the correct solution. The answer the question is The student made an error in Step 1.

7 0

2 years ago

If minor arc AB measures 9 inches, what is the length of the radius of circle C? If necessary, round your answer to the nearest

Art [367]

Complete Question

If minor arc AB measures 9 inches, what is the length of the radius of circle C? Where the radians is 0.75 radians

If necessary, round your answer to the nearest inch. 6 inches 12 inches 18 inches 24 inches

Answer:

12 inches

Step-by-step explanation:

The formula for Arc length =

Arc length = r θ

θ = 0.75

Arc length = 9 inches

Hence:

r = 9/ θ

r = 9/0.75

r = 12 inches

3 0

2 years ago

Read 2 more answers

Find the range of the following details

irga5000 [103]

Mean- 6.22
Median- 5
Mode- 11 (appears 2 times)
Range- 14

5 0

2 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

2 years ago

Which expression below gives the average rate of change of the function g(x) = -x2 - 4x on the interval 6 ≤ x ≤ 8 ?

Paha777 [63]

Answer: -18

<u>Explanation:</u>

Average rate of change is slope: $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ . Need to find the y-coordinates:

g(x) = -x² - 4x

g(6) = -(6)² - 4(6)

= -36 - 24

= -60

⇒ (6, -60)

g(8) = -(8)² - 4(8)

= -64 - 32

= -96

⇒ (8, -96)

m = $\frac{-96 - (-60)}{8 - 6}$

= $\frac{-36}{2}$

= -18

7 0

2 years ago

Read 2 more answers