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4 years ago

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

Mathematics

1 answer:

sergejj [24]4 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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3x -9 =7x + 12<br>Find X

Lorico [155]

Simple....

you have:

3x-9=7x+12

To even solve this....isolate the variable your are trying to solve for....

3x-9=7x+12
-3x -3x

-9=4x+12

-9=4x+12
-12 -12

-21=4x

Simplify.....

$\frac{-21}{4} = \frac{4x}{4}$

$\frac{-21}{4} =x$

Thus, your answer.

5 0

3 years ago

Can someone help? 20 points :D

love history [14]

Q should equal -4.2.

8 0

3 years ago

Read 2 more answers

If X is a r.v. such that E(X^n)=n! Find the m.g.f. of X,Mx(t). Also find the ch.f. of X,and from this deduce the distribution of

astraxan [27]

$M_X(t)=\mathbb E(e^{Xt})$
$M_X(t)=\mathbb E\left(1+Xt+\dfrac{t^2}{2!}X^2+\dfrac{t^3}{3!}X^3+\cdots\right)$
$M_X(t)=\mathbb E(1)+t\mathbb E(X)+\dfrac{t^2}{2!}\mathbb E(X^2)+\dfrac{t^3}{3!}\mathbb E(X^3)+\cdots$
$M_X(t)=1+t+t^2+t^3+\cdots$
$M_X(t)=\displaystyle\sum_{k\ge0}t^k=\frac1{1-t}$

provided that $|t|$ .

Similarly,

$\varphi_X(t)=\mathbb E(e^{iXt})$
$\varphi_X(t)=1+it+(it)^2+(it)^3+\cdots$
$\varphi_X(t)=(1-t^2+t^4-t^6+\cdots)+it(1-t^2+t^4-t^6+\cdots)$
$\varphi_X(t)=(1+it)(1-t^2+t^4-t^6+\cdots)$
$\varphi_X(t)=\dfrac{1+it}{1+t^2}=\dfrac1{1-it}$

You can find the CDF/PDF using any of the various inversion formulas. One way would be to compute

$F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{e^{itx}\varphi_X(-t)-e^{-itx}\varphi_X(t)}{it}\,\mathrm dt$

The integral can be rewritten as

$\displaystyle\int_0^\infty\frac{2i\sin(tx)-2it\cos(tx)}{it(1+t^2)}\,\mathrm dt$

so that

$F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt$

There are lots of ways to compute this integral. For instance, you can take the Laplace transform with respect to $x$ , which gives

$\displaystyle\mathcal L_s\left\{\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt\right\}=\int_0^\infty\frac{1-s}{(1+t^2)(s^2+t^2)}\,\mathrm dt$
$=\displaystyle\frac{\pi(1-s)}{2s(1+s)}$

and taking the inverse transform returns

$F_X(x)=\dfrac12+\dfrac1\pi\left(\dfrac\pi2-\pi e^{-x}\right)=1-e^{-x}$

which describes an exponential distribution with parameter $\lambda=1$ .

6 0

3 years ago

1. Write an expression equivalent to the following:

Vika [28.1K]

A. x+5 (divided by 4)
b. 2x+20 (multiplied by 2)

7 0

3 years ago

Find the length of AP in<br>the diagram below,<br>rounding to the nearest tenth

tester [92]

Answer:

6.9 to the nearest tenth.

Step-by-step explanation:

Use the external tangent - chord theorem:

AP^2 = CP * BP

AP^2 = 12 * 4 = 48

AP = √48

= 6.928

4 0

4 years ago