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2 years ago

Can someone explain to me in words how to do this problem 3x+122=22x-11

Mathematics

1 answer:

inn [45]2 years ago

8 0

When you have this type of problem, you need to combine the like-terms and isolate the variable.

3x + 122 = 22x - 11

Add 11 to both sides to get rid of it

3x + 122 + 11 = 22x - 11 + 11 (-11 + 11=0)

3x + 133 = 22x

Then you would bring the 3x to the other side, so subtract 3x from both sides

3x + 133 = 22x

-3x -3x

133 = 22x - 3x

133 = 19x

Then divide both sides by 19 to isolate x

133/19 = 19x/19

133/19 = 7, so x = 7

Hope this helps!!

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2 years ago

Find the taylor series for f(x) centered at the given value of a. [assume that f has a power series expansion. do not show that

RUDIKE [14]

The taylor series for the f(x)=8/x centered at the given value of a=-4 is -2+2(x+4)/1!-24/16 $(x+4)^{2}$ /2!+...........

Given a function f(x)=9/x,a=-4.

We are required to find the taylor series for the function f(x)=8/x centered at the given value of a and a=-4.

The taylor series of a function f(x)= $f(a)+f^{1}(a)(x-a)/1!+ f^{11}(a)(x-a)^{2} /2! +f^{111}(a)(x-a)a^{3}/3!+..........$

Where the terms in f prime $f^{1}$ (a) represent the derivatives of x valued at a.

For the given function.f(x)=8/x and a=-4.

So,f(a)=f(-4)=8/(-4)=-2.

$f^{1}$ (a)= $f^{1}$ (-4)=-8/( $-4)^{2}$

=-8/16

=-1/2

The series of f(x) is as under:

f(x)=f(-4)+ $f^{1}(-4)(x+4)/1!+ f^{11}(-4)(x+4)^{2}/2!.............$

$=8/(-4)-8/(-4)^{2} (-4)(x+4)/1!+ 24/(-4)^{3} (-4)(x+4)^{2}/2!.............$

=-2+2(x+4)/1!-24/16 $(x+4)^{2}$ /2!+...........

Hence the taylor series for the f(x)=8/x centered at the given value of a=-4 is -2+2(x+4)/1!-24/16 $(x+4)^{2}$ /2!+...........

Learn more about taylor series at brainly.com/question/23334489

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11 months ago

Which system of inequalities has a solution set that is a line?

Korolek [52]

We're going to solve each of the systems to determine the solution.

System $1$

$x+y \geq 3\\ x+y \leq 3$

using a graph tool

the solution is the line $x+y=3$

see the attached figure N $1$

System $2$

$x+y \geq -3\\ x+y \leq 3$

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the solution is the shaded area

see the attached figure N $2$

System $3$

$x+y > 3\\ x+y < 3$

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The system has no solution

see the attached figure N $3$

System $4$

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the solution is the shaded area

see the attached figure N $4$

therefore

the answer is

System $1$ 

$x+y \geq 3\\ x+y \leq 3$

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3 years ago

Read 2 more answers