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BaLLatris [955]

3 years ago

11

4x - 6+ 2x - 4= 2 + 3x

2 answers:

Masteriza [31]3 years ago

8 0

Answer:

4x-6+2x-4=2+3x

4x+2x-3x=2+6+4

6x-3x=12

3x/3=12/3

x=4

Salsk061 [2.6K]3 years ago

3 0

Answer:

$4x - 6 + 2x - 4 = 2 + 3x \\ 4x + 2x - 3x = 2 + 6 + 4 \\ 3x = 12 \\ \\ x = \frac{12}{3} \\ \\ x = 4$

I hope I helped you^_^

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Solve for y<br> 2(3 + 3y) + y = 11

garik1379 [7]

<h2><u>EQUATION</u></h2><h3>Exercise</h3>

2(3 + 3y) + y = 11

First, apply the distributive property:

2(3 + 3y) + y = 11

6 + 6y + y = 11

6 + 7y = 11

Substract 6 from both sides:

6 - 6 + 7y = 11 - 6

7y = 5

Divide both sides by 7:

$\dfrac{7y}{7} = \dfrac{5}{7}$

$\boxed{y = \dfrac{5}{7}}$

<h3><u>Answer</u>. The value of y = 5/7.</h3>

8 0

3 years ago

What is the missing term in the quadratic expression below?<br><br> (2x-3)(x-4)=2x^2+____ -12

cupoosta [38]

Answer:

ok

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Dahasolnce [82]

Answer:

a) $x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$ , $x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$ , b) $x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$ , $x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$

Step-by-step explanation:

a) The equation must be rearranged into a form with one fundamental trigonometric function first:

$\sqrt{3}\cdot \csc x - 2 = 0$

$\sqrt{3} \cdot \left(\frac{1}{\sin x} \right) - 2 = 0$

$\sqrt{3} - 2\cdot \sin x = 0$

$\sin x = \frac{\sqrt{3}}{2}$

$x = \sin^{-1} \frac{\sqrt{3}}{2}$

Value of x is contained in the following sets of solutions:

$x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$

$x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$

b) The equation must be simplified first:

$\cos x + 1 = - \cos x$

$2\cdot \cos x = -1$

$\cos x = -\frac{1}{2}$

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

Value of x is contained in the following sets of solutions:

$x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$

$x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$

7 0

3 years ago

Use the diagram to find the measure of angle KHL.

bija089 [108]

Answer:

m∠KHL = 43°

Step-by-step explanation:

From the picture attached,

m∠KHL ≅ m∠GHL [Given in the picture]

Now substituting the values of the angles,

(3x + 1) = (5x - 27)

1 + 27 = 5x - 3x

28 = 2x

14 = x

m∠KHL = (3x + 1)° = (3 × 14) + 1

= 42 + 1

= 43°

Therefore, measure of ∠KHL = 43°

3 0

3 years ago

What is the center and radius of the following circle (x - 5) ^ 2 + (y + 2) ^ 2 = 16

murzikaleks [220]

Answer:

Step-by-step explanation:

if center=(h,k)

radius=r

eq. of circle is (x-h)^2+(y-k)^2=r^2

given eq. is (x-5)^2+(y+2)^2=16=4^2

comparing

center =(5,-2)

radius r=4

4 0

3 years ago