All categories

3 years ago

Solve the following equation. show all your work x/x-2+x-1/x+1=-1

Mathematics

2 answers:

Andru [333]3 years ago

6 0

Answer:

X=1

Step-by-step explanation:

dmitriy555 [2]3 years ago

3 0

We hates fractions, yess precious, nasty fracitons

multiply both sides by (x-2)(x+1) to clea fractions
x(x+1)+(x-1)(x-2)=(-1)(x-2)(x+1)
distribute
x^2+x+x^2-3x+2=-x^2+x-2
add like terms
2x^2-2x+2=-x^2+x+2
add x^2 to both sdes
3x^2-2x+2=x+2
minus x from both sides
3x^2-3x+2=2
minus 2 both sides
3x^2-3x=0
factor
(3x)(x-1)=0
set to zero

3x=0
x=0

x-1=0
x=1

x=0 or 1

You might be interested in

Find the rate of change in the table

mixas84 [53]

Answer:

I'm pretty sure it's +(-6).

Step-by-step explanation:

Basically for these, you find the slope. The slope equation is y1-y2/x1-x2.

Pick two points.

I'll just do (1,15) and (2,9).

15-9/1-2 = 6/-1 = -6.

I don't know if you were taught it this way, but my math teacher always told us that the rate of change had to be positive.

So you'd say the rate of change is +(-6), not -6.

Hope that sorta helped lol

3 0

3 years ago

To graph the equation 2x + 5y = 10, Zeplyn draws a line through the points (5, 0) and (0, What is the slope of the line represen

wariber [46]

Answer: -2/5

Step-by-step explanation:

To find the slope, we set the function use y = mx + b forms.

2x + 5y = 10

5y = -2x + 10 Minus 2x in both side

y = -2/5x + 10/5 Divide 5 from both side

y = -2/5x + 2

In the y = mx + b, the m before variable x is always the slope, so -2/5 is the slope for 2x + 5y = 10.

8 0

3 years ago

An article suggests that a poisson process can be used to represent the occurrence of structural loads over time. suppose the me

kirill115 [55]

Answer:

a) $\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

b) $P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

c) $e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}$

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution"

Solution to the problem

Let X our random variable who represent the "occurrence of structural loads over time"

For this case we have the value for the mean given $\mu = 0.5$ and we can solve for the parameter $\lambda$ like this:

$\frac{1}{\lambda} = 0.5$

$\lambda =2$

So then $X(t) \sim Poi (\lambda t)$

X follows a Poisson process

Part a

For this case since we are interested in the number of loads in a 2 year period the new rate would be given by:

$\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

Part b

For this case we want the following probability:

$P(X(2) >6)$

And we can use the complement rule like this

$P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

And we can solve this like this using the masss function:

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

Part c

For this case we know that the arrival time follows an exponential distribution and let T the random variable:

$T \sim Exp(\lambda=2)$

The probability of no arrival during a period of duration t is given by:

$f(T) = e^{-\lambda t}$

And we want to find a value of t who satisfy this:

$e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

3 0

3 years ago

Geometry 100 pts plz help!!! screenshot attatched

nirvana33 [79]

Answer:

1. A

2. D

3. C

4. E

5. B

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Help me plz with this question

Dominik [7]

Answer:

y= -6x+23

Step-by-step explanation:

perpendicular: opposite signs and reciprocal

m= 1/6 and y-intercept = 23

y= -6x+23

3 0

3 years ago