<img src="https://tex.z-dn.net/?f=4x%2B2%3D2%28x%2B6%29" id="TexFormula1" title="4x+2=2(x+6)" alt="4x+2=2(x+6)" align="absmiddle

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solmaris [256]

3 years ago

" class="latex-formula">

Solve for x.

Mathematics

2 answers:

IgorLugansk [536]3 years ago

7 0

Answer:

$x = 5$

Step-by-step explanation:

Step 1: Distribute

$4x + 2 = 2(x + 6)$

$4x + 2 = (2 * x) + (2 * 6)$

$4x + 2 = 2x + 12$

Step 2: Subtract 2 from both sides

$4x + 2 - 2 = 2x + 12 - 2$

$4x = 2x + 10$

Step 3: Subtract 2x from both sides

$4x - 2x = 2x - 2x + 10$

$2x = 10$

Step 4: Divide both sides by 2

$2x / 2 = 10 / 2$

$x = 5$

Answer: $x = 5$

djyliett [7]3 years ago

3 0

Hi there!

»»————-　★　————-««

I believe your answer is:

$x = 5$

»»————-　★　————-««

Here’s why:

⸻⸻⸻⸻

$\boxed{\text{Solving for 'x'...}}\\\\4x + 2 =2(x+6)\\------------\\\rightarrow 4x + 2 = 2x + 12\\\\\rightarrow4x - 2x +2 = 2x - 2x + 12\\\\\rightarrow 2x + 2 = 12\\\\\rightarrow2x + 2 - 2 = 12 - 2\\\\\rightarrow 2x = 10\\\\\rightarrow \frac{2x=10}{2}\\\\\rightarrow \boxed{x = 5}$

⸻⸻⸻⸻

»»————-　★　————-««

Hope this helps you. I apologize if it’s incorrect.

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In a recent year, an author wrote 171 checks. Use the Poisson distribution to find the probability that, on a randomly selecte

sveticcg [70]

Answer: 0.3741

Step-by-step explanation:

Poison probability ;

P(x) = [U(^x) e(^-U)] ÷ x!

Where U = mean

Note: e = exponential symbol

Number of checks that year = 171

Number of days in a year = 365

U = 171/365 = 0.468

Average checks per day = 0.4685

Probability that at least one check was written per day is can be calculated by;

P(not 0) = 1 - P(0)

Therefore,

P(x) = [U(^x) e(^-U)] ÷ x!

P(0) = [ 0.4685^0 * e^-0.4685] ÷ 0!

P(0) = [ 1 * 0.6259] ÷ 1

P(0) = 0.6259

Therefore,

P(not 0) = 1 - 0.6259 = 0.3741

4 0

3 years ago

Find the limit............

liubo4ka [24]

Answer: $-\frac{9}{x^2}$

====================================================

Work Shown:

$\displaystyle \lim_{\Delta x \to 0^{+}} \frac{\frac{9}{x+\Delta x} - \frac{9}{x}}{\Delta x}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{\frac{9x}{x(x+\Delta x)} - \frac{9(x+\Delta x)}{x(x+\Delta x)}}{\Delta x}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{\frac{9x-9(x+\Delta x)}{x(x+\Delta x)}}{\Delta x}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{9x-9x-9\Delta x}{x\Delta x(x+\Delta x)}\\\\\\$

$\displaystyle \lim_{\Delta x \to 0^{+}} \frac{-9\Delta x}{x\Delta x(x+\Delta x)}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{-9}{x(x+\Delta x)}\\\\\\\displaystyle \frac{-9}{x(x+0)}\\\\\\\displaystyle -\frac{9}{x^2}\\\\\\$

The trick here is to first simplify the expression so that the $\Delta x$ term in the very lower denominator cancels out. This is to avoid dividing by zero. Once this division by zero error is avoided, we can replace the delta x term with 0 and evaluate the limit as $\Delta x$ approaches 0 from the right or positive direction.