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3 years ago

Which would not be a step in solving 3x + 1 + 2x = 2+ 4x?

1 answer:

6 0

Well here is how you solve.

You first combine like-terms.
3x + 1 + 2x = 2 + 4x
5x + 1 = 2 + 4x
Then you isolate the variable by subtracting 1 from both sides.
5x = 1 + 4x
Then you do the same with 4x.
1x = 1
Now divide.
x = 1
Now I know this isn't the exact answer but if you look at the options you have, one of those shouldn't be shown, then that would be your answer. I hope this helps love! :)

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Please help 6x + 32 = -2x

netineya [11]

Answer: the answer is x=-4.

Step-by-step explanation:

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1 year ago

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I hope my answer has come to your help. God bless you and have a nice day ahead!

5 0

3 years ago

Deandre just bought 9 bags of 15 cookies each. He already had 6 cookies in a jar. How many cookies does deandre have now?

djverab [1.8K]

Answer:

141 cookies

Step-by-step explanation:

amount of cookies

= 9(15) + 6

= 135 + 6

= 141

6 0

3 years ago

Find the length of the missing

Ivahew [28]

Answer:

Step-by-step explanation:

using pythagoras theorem

here 15 is hypotenuse since it is opposite 0f 90 degree

9 and x are the other smaller sides of a triangle

according to pythagoras thorem the sum of square of two smaller sides of a triangle is equal to the square of hypotenuse. So,

9^2 + x^2 = 15^2

81 + x^2 = 225

x^2 = 225 - 81

x^2 = 144

x = $\sqrt{144$

x = 12

5 0

3 years ago

Read 2 more answers

Factor completely. <img src="https://tex.z-dn.net/?f=x%5E%7B8%7D-%5Cfrac%7B1%7D%7B81%7D" id="TexFormula1" title="x^{8}-\fra

Eduardwww [97]

We have 3⁴ = 81, so we can factorize this as a difference of squares twice:

$x^8 - \dfrac1{81} = \left(x^2\right)^4 - \left(\dfrac13\right)^4 \\\\ x^8 - \dfrac1{81} = \left(\left(x^2\right)^2 - \left(\dfrac13\right)^2\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)$

Depending on the precise definition of "completely" in this context, you can go a bit further and factorize $x^2-\frac13$ as yet another difference of squares:

$x^2 - \dfrac13 = x^2 - \left(\dfrac1{\sqrt3}\right)^2 = \left(x-\dfrac1{\sqrt3}\right)\left(x+\dfrac1{\sqrt3}\right)$

And if you're working over the field of complex numbers, you can go even further. For instance,

$x^4 + \dfrac19 = \left(x^2\right)^2 - \left(i\dfrac13\right)^2 = \left(x^2 - i\dfrac13\right) \left(x^2 + i\dfrac13\right)$

But I think you'd be fine stopping at the first result,

$x^8 - \dfrac1{81} = \boxed{\left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)}$

6 0

3 years ago