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3 years ago

How do you solve 8|x+2|-6=5|x+2|+3

Mathematics

1 answer:

sergejj [24]3 years ago

7 0

Answer:

x = 1 or x = -5

Step-by-step explanation:

Solve for x over the integers:

8 abs(x + 2) - 6 = 5 abs(x + 2) + 3

Subtract 5 abs(x + 2) - 6 from both sides:

3 abs(x + 2) = 9

Divide both sides by 3:

abs(x + 2) = 3

Split the equation into two possible cases:

x + 2 = 3 or x + 2 = -3

Subtract 2 from both sides:

x = 1 or x + 2 = -3

Subtract 2 from both sides:

Answer: x = 1 or x = -5

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Evaluate 2(N + 3) if n = 5 show your work

algol [13]

ANSWER:

N = 16

STEP BY STEP EXPLANATION:

2(5+3)

n=16

6 0

3 years ago

Can segments with lengths of 15, 20, and 36 form a triangle?

strojnjashka [21]

Ok, so remember

the legnth of the longest side must be LESS THAN the sum of the measures of the other 2 sides
if no longest side (becasue it has 2 longest sides or 3 equal sides), then it can form a triangle

so the longest side is 36

36 must be less than the sum of 15 and 20
36<15+20
36<35
false

therfor it cannot form a triangle
try it yourself, you cannot connect them that way

5 0

3 years ago

5/8 - 3/4(8 - 1/3) + 1

Anon25 [30]

The answer is definitely -4 1/8

6 0

3 years ago

Read 2 more answers

The joint probability density function of X and Y is given by fX,Y (x, y) = ( 6 7 x 2 + xy 2 if 0 < x < 1, 0 < y < 2

fredd [130]

I'm going to assume the joint density function is

$f_{X,Y}(x,y)=\begin{cases}\frac67(x^2+\frac{xy}2\right)&\text{for }0$

a. In order for $f_{X,Y}$ to be a proper probability density function, the integral over its support must be 1.

$\displaystyle\int_0^2\int_0^1\frac67\left(x^2+\frac{xy}2\right)\,\mathrm dx\,\mathrm dy=\frac67\int_0^2\left(\frac13+\frac y4\right)\,\mathrm dy=1$

b. You get the marginal density $f_X$ by integrating the joint density over all possible values of $Y$ :

$f_X(x)=\displaystyle\int_0^2f_{X,Y}(x,y)\,\mathrm dy=\boxed{\begin{cases}\frac67(2x^2+x)&\text{for }0$

c. We have

$P(X>Y)=\displaystyle\int_0^1\int_0^xf_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx=\int_0^1\frac{15}{14}x^3\,\mathrm dx=\boxed{\frac{15}{56}}$

d. We have

$\displaystyle P\left(X$

and by definition of conditional probability,

$P\left(Y>\dfrac12\mid X\frac12\text{ and }X$

$\displaystyle=\dfrac{28}5\int_{1/2}^2\int_0^{1/2}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\boxed{\frac{69}{80}}$

e. We can find the expectation of $X$ using the marginal distribution found earlier.

$E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}$

f. This part is cut off, but if you're supposed to find the expectation of $Y$ , there are several ways to do so.

Compute the marginal density of $Y$ , then directly compute the expected value.

$f_Y(y)=\displaystyle\int_0^1f_{X,Y}(x,y)\,\mathrm dx=\begin{cases}\frac1{14}(4+3y)&\text{for }0$

$\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87$

Compute the conditional density of $Y$ given $X=x$ , then use the law of total expectation.

$f_{Y\mid X}(y\mid x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\begin{cases}\frac{2x+y}{4x+2}&\text{for }0$

The law of total expectation says

$E[Y]=E[E[Y\mid X]]$

We have

$E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}$

$\implies E[Y\mid X]=1+\dfrac1{6X+3}$

This random variable is undefined only when $X=-\frac12$ which is outside the support of $f_X$ , so we have

$E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87$

5 0

3 years ago

Solve: 2x+6-10x=30

LiRa [457]

The answer will be -3

3 0

3 years ago