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3 years ago

2(x-3) +9=3(x+1) + x

Mathematics

1 answer:

Licemer1 [7]3 years ago

5 0

Answer:

x = 0

Step-by-step explanation:

2(x - 3) + 9 = 3(x + 1) + x

2x - 6 + 9 = 3x + 3 + x

2x + 3 = 4x + 3

2x - 4x = 3 - 3

-2x = 0

2x = 0

x = 0/2

x = 0

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B9%7D" id="TexFormula1" title="\frac{1}{9}" alt="\frac{1}{9}" align="absmiddle

Jobisdone [24]

$\dfrac{1}{9} - \dfrac{5}{6}$

Set up

$\dfrac{2}{18} - \dfrac{15}{18}$

Get all fractions to a common denominator which will allow you to subtract them

$\dfrac{2 - 15}{18}$

Subtract numerators

$\boxed{- \dfrac{13}{18}}$

Simplify

4 0

3 years ago

Convert from scientific notation to standard form 4.83 x 10^-3

ANTONII [103]

4.83 x 10^-3 can be re-written in standard form by moving the decimal point 3 units to the left: 0.00483.

Note: If you want to be mathematically correct, please write 10^(-3), not 10^-3. If the exponent is +, you don't need the parentheses.

5 0

3 years ago

What is 4and1/2 subtract 1/4

umka21 [38]

4 1/2=4 2/4

4 2/4-1/4=4 1/4 or 4.25

8 0

3 years ago

F(x)=x-1/x^2-x-6 which is the graph of

hoa [83]

Answer:

The graph is attached below.

Step-by-step explanation:

<em>As you have not added the graph, so I will be solving the function for a graph.</em>

Given the function

$f\left(x\right)=x-\frac{1}{x^2}-x-6$

$x-\mathrm{axis\:interception\:points\:of\:}-\frac{1}{x^2}-6:$

$\mathrm{x-intercept\:is\:a\:point\:on\:the\:graph\:where\:}y=0$

$-\frac{1}{x^2}-6=0$

$-1-6x^2=0$

$\mathrm{No\:Solution\:for}\:x\in \mathbb{R}$

$\mathrm{No\:x-axis\:interception\:points}$

$y-\mathrm{axis\:interception\:point\:of\:}-\frac{1}{x^2}-6:$

$y\mathrm{-intercept\:is\:the\:point\:on\:the\:graph\:where\:}x=0$

As we know that the domain of a function is the set of input or argument values for which the function is real and defined.

$\mathrm{Domain\:of\:}\:-\frac{1}{x^2}-6\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x0\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:0\right)\cup \left(0,\:\infty \:\right)\end{bmatrix}$

$\mathrm{Since}\:x=0\:\mathrm{is\:not\:in\:domain}$

$\mathrm{No\:y-axis\:interception\:point}$

$\mathrm{Asymptotes\:of}\:-\frac{1}{x^2}-6:\quad \mathrm{Vertical}:\:x=0,\:\mathrm{Horizontal}:\:y=-6$

$\mathrm{Range\:of\:}-\frac{1}{x^2}-6:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)$

The graph is attached below.

5 0

3 years ago

Can someone please tell me what to do for this?? Thank you :))

bezimeni [28]

take it and leave it somewhere u will forget about it and move on with ur life lol

5 0

3 years ago