Can someone solve this and show your work <br><br> g=6x<br> Solve for X

Write an expression in simplest form for the perimeter of the exercise mat

Verizon [17]

The simplest expression is (2w)+(2*3w), because to calculate perimeter, you have to use this formula: 2length+2width.

6 0

3 years ago

Write the inequality represents the verbal expression-<br> All real numbers less than 69

Lorico [155]

Let be M the set of the real numbers less thant 69

so we write it as follow M = { x∈R / x ∠ 69}

7 0

3 years ago

which statement is true about the equations -3x+4y=12 and 1 over 4 x - 1 over 3 y =1 A) The system of the equations has exactly

Citrus2011 [14]

Hi again :)

We need to solve -3x+4y=12 for x

Let's start by adding -4y to both sides

-3x+4y-4y=12-4y

-3x=-4y+12

x = (-4y+12)/-3

x= 4/3 y -4

Now substitute 4/3 y -4 for x in 1/4 x - 1/3 y =1

1/4 x -1/3 y =1

1/4 (4/3 y -4) -1/3 y =1

Use the distributive property

(1/4)(4/3 y) + (1/4)(-4) -1/3 y =1

1/3 y -1 - 1/3 y =1

Now combine like terms

(1/3y -1/3y) + (-1) =1

= -1

-1 = 1

Now add 1 to both sides

0=2

So there are no Solutions

The answer is C

I hope that's help Will :)

6 0

3 years ago

Read 2 more answers

Solve the equation equations by using completing the square method

Anna11 [10]

Answer:

Step-by-step explanation:

(a) 4x2 -- 3x - 2 = 0

x1,2 = -b ± √b²- 4ac

---------------------

2a

x1,2 = 3 ±√41

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8

(b) (2x - 3)2 = 6

4x² -12x + 9 -6 = 0

4x² - 12 x + 3 = 0

Step-by-step explanation:

(a) 4x2 -- 3x - 2 = 0

x1,2 = -b ± √b²- 4ac

---------------------

2a

x1,2 = 3 ±√41

--------------

8

(b) (2x - 3)2 = 6

4x² -12x + 9 -6 = 0

4x² - 12 x + 3 = 0

5 0

3 years ago

Plźzzz help me ... I'll give brainliest

fgiga [73]

Answer:

$\displaystyle x = \frac{1}{3}$ .

Step-by-step explanation:

Consider the double angle identity for tangents:

$\displaystyle \tan(2\theta) = \frac{2\tan{\theta}}{1 - \tan^{2}{\theta}}$ .

Also, for two complementary angles,

$\displaystyle \tan{\left(\frac{\pi}{2} - \theta \right)} = \frac{1}{\tan{\theta}}$ .

Subtract $\tan^{-1}(x + 1)$ from both sides of this equation:

$\displaystyle 2\tan^{-1}{x} = \frac{\pi}{2} + (-\tan^{-1}(x + 1))$ .

Take the tangent of both sides of this equation:

$\displaystyle \tan(2\tan^{-1}{x}) = \tan\left(\frac{\pi}{2} -\tan^{-1}\left(x + 1\right)\right)$ .

Apply the double-angle identity to the left-hand side of this equation:

$\displaystyle \tan(2\tan^{-1}{x}) \implies \frac{2\tan(\tan^{-1}x)}{1 - \tan^{2}(\tan^{-1}x)}\implies \frac{2x}{1 - x^{2}}$ .

The two angles $\displaystyle \left(\frac{\pi}{2} + (-\tan^{-1}\left(x + 1\right))\right)$ and $(x - 1)$ are complementary. Therefore, for the right-hand side of this equation,

$\displaystyle \tan\left(\frac{\pi}{2} -\tan^{-1}\left(x + 1\right)\right)\implies \frac{1}{\tan(\tan^{-1}(x + 1))} \implies \frac{1}{x + 1}$ .

Equate the two sides of this equation:

$\begin{aligned} & \frac{2x}{1 - x^{2}} = \frac{1}{x + 1}\\ \implies & \frac{2x}{(1 - x)(1 + x)} = \frac{1}{1 + x}\\\implies & 2x = 1 - x,\; x \ne 1,\; x \ne -1\\\implies & x = \frac{1}{3}\end{aligned}$ .

5 0

3 years ago

Can someone solve this and show your work g=6x Solve for X