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2 years ago

-12+5x=15-4x Please help me :( I need to know how to get the answer as well

Mathematics

2 answers:

tankabanditka [31]2 years ago

5 0

Answer:

We simplify the equation to the form, which is simple to understand

5x+15=4x-12

We move all terms containing x to the left and all other terms to the right.

+5x-4x=-12-15

We simplify left and right side of the equation.

+1x=-27

then divide both sides of the equation by 1 to get x.

x=-27

Oduvanchick [21]2 years ago

5 0

Answer: 3

Step-by-step explanation:

-12+5x=15-4x

5x+4x

-12+9x+15

15+12

9x/27

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Anika [276]

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Step-by-step explanation:

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2 years ago

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MATH HELP!!! 100PTS!!!

Lostsunrise [7]

Answer:

$x =\dfrac{20 +\sqrt{454}}{18}, \quad \dfrac{20 -\sqrt{454}}{18}$

Step-by-step explanation:

<u>Given functions</u>:

$p(x)=2x-4$

$r(x)=\dfrac{6x-1}{9x+1}$

Solve for p(x) = r(x):

$\begin{aligned}p(x) & = r(x)\\\implies 2x-4 & = \dfrac{6x-1}{9x+1}\\(2x-4)(9x+1)&=6x-1\\18x^2+2x-36x-4&=6x-1\\18x^2-40x-3&=0\end{aligned}$

As the found quadratic equation cannot be factored, use the Quadratic Formula to solve for x:

<u>Quadratic Formula</u>

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Therefore:

$a=18, \quad b=-40, \quad c=-3$

Substitute the values of a, b and c into the <u>quadratic formula</u> and solve for x:

$\begin{aligned}\implies x & =\dfrac{-(-40) \pm \sqrt{(-40)^2-4(18)(-3)} }{2(18)}\\\\& =\dfrac{40 \pm \sqrt{1816}}{36}\\\\& =\dfrac{40 \pm \sqrt{4 \cdot 454}}{36}\\\\& =\dfrac{40 \pm \sqrt{4}\sqrt{454}}{36}\\\\& =\dfrac{40 \pm2\sqrt{454}}{36}\\\\& =\dfrac{20 \pm\sqrt{454}}{18}\end{aligned}$

Therefore, the solutions are:

$x =\dfrac{20 +\sqrt{454}}{18}, \quad \dfrac{20 -\sqrt{454}}{18}$

Learn more about quadratic equations here:

brainly.com/question/27750885

brainly.com/question/27739892

6 0

1 year ago

Please help with number 6 ASAP I will mark branlist

DerKrebs [107]

Answer:

<h2>12</h2><h2>Step-by-step explanation:</h2>

$4\sqrt{9}\\\\\mathrm{Factor\:the\:number:\:}\:9=3^2\\=4\sqrt{3^2}\\\\\mathrm{Apply\:radical\:rule}:\quad \sqrt{a^2}=a,\:\quad \:a\ge 0\\\sqrt{3^2}=3\\\\=4\times\:3\\\\=12$