The diagram shows AKLM. Which term describes point N?
L
N
M

The correct diagnosis code to report a tibial closed fracture, a proximal end, of the left leg, the initial encounter is A.S82.191A.

Explanation:

Diagnosis codes are used to accurately report a patient's condition or injury. The code S82.191A specifically refers to a closed fracture of the proximal end of the tibia on the left leg, with this being the initial encounter for treatment.

The other options, S82.191B, S82.102A, and S82.102B do not accurately describe the condition or injury in question. S82.191B refers to a subsequent encounter for the same injury, while S82.102A and S82.102B refer to a closed fracture of the right leg's upper end of the tibia.

Therefore, the correct answer is A.S82.191A.

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Value of limit expression

[tex]\lim_{\theta \to 0} \dfrac{10\theta}{3sin\theta}[/tex]  [tex]=\dfrac{5}{9}[/tex]

Value of limit expression

[tex]\lim_{x \to 1} \dfrac{sin(x -1)}{x^{2} + 5x - 6}[/tex] [tex]= \dfrac{1}{7}[/tex]

In mathematics, the limit is the value that the function approaches as the input approaches the value. Limits are essential in calculus and mathematical analysis and are used to define continuity, derivatives and integrals  

Given

[tex]a) \lim_{\theta \to 0} \dfrac{10\theta}{3sin\theta}[/tex]

[tex]\dfrac{10}{3} \lim_{\theta \to 0} \dfrac{\theta}{sin6\theta}[/tex]

[tex]sin\theta = \theta - \dfrac{\theta^{3}}{3!} + \dfrac{\theta^{5}}{5!} + .......[/tex]

[tex]\dfrac{10}{3} \lim_{\theta \to 0} \dfrac{\theta}{6\theta - \dfrac{(6\theta)^{3} }{3!} +\dfrac{(6\theta)^{5} }{5!}- ......}[/tex]

[tex]\dfrac{10}{3} \lim_{\theta \to 0} \dfrac{\theta}{(6\theta)(1 - \dfrac{(6\theta)^{2} }{3!} +\dfrac{(6\theta)^{4} }{5!}- ......)}[/tex][tex]\dfrac{10}{3\times6} \lim_{\theta \to 0} \dfrac{\theta}{(\theta)(1 - \dfrac{(6\theta)^{2} }{3!} +\dfrac{(6\theta)^{4} }{5!}- ......)}[/tex][tex]\dfrac{10}{3\times6} \lim_{\theta \to 0} \dfrac{1}{(1 - \dfrac{(6\theta)^{2} }{3!} +\dfrac{(6\theta)^{4} }{5!}- ......)}[/tex]

[tex]\dfrac{10}{3\times6} \times \dfrac{1}{(1 - \dfrac{(0)^{2} }{3!} +\dfrac{(0)^{4} }{5!}- ......)}[/tex]

[tex]=\dfrac{5}{9}[/tex]

[tex]b)\lim_{x \to 1} \dfrac{sin(x -1)}{x^{2} + 5x - 6}[/tex]

[tex]\lim_{x \to 1} \dfrac{(x -1) - \dfrac{(x - 1)^{3}}{3!}-\dfrac{(x - 1)^{5}}{5!} - ......}{x^{2} + 6x - x -6}[/tex][tex]\lim_{x \to 1} \dfrac{(x -1) - \dfrac{(x - 1)^{3}}{3!}-\dfrac{(x - 1)^{5}}{5!} - ......}{x(x + 6) - 1(x +6)}[/tex][tex]\lim_{x \to 1} \dfrac{(x - 1)(1 - \dfrac{(x - 1)^{2}}{3!}-\dfrac{(x - 1)^{4}}{5!} - ......)}{(x- 1)(x +6)}[/tex]

[tex]\dfrac{(1 - \dfrac{(1 - 1)^{2}}{3!}-\dfrac{(1 - 1)^{4}}{5!} - ......)}{(1 +6)}[/tex]

[tex]=\dfrac{1}{7}[/tex]

Hence, [tex]\dfrac{5}{9}[/tex] is value of limit expression [tex]\lim_{\theta \to 0} \dfrac{10\theta}{3sin\theta}[/tex]

and [tex]\dfrac{1}{7}[/tex] is value of  expression [tex]\lim_{x \to 1} \dfrac{sin(x -1)}{x^{2} + 5x - 6}[/tex]

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The value of the functions at the limit value of the input variable using the piecewise function graph are;

[tex]\lim\limits_{x\to2^+}\frac{f(x)-1}{f(x + 4)} =\underline{ \frac{1}{6}}[/tex]

[tex]\lim\limits_{x\to 1^-}f(f(x) +1) = \underline{5}[/tex]

[tex]\lim\limits_{h\to0}\frac{f(6+h)-f(6)}{h} = \underline{2}[/tex]

The limit of a function is the value of the function as the input value approaches the limit.

The graph of the piecewise function indicates that at x → 2⁺ is; f(x) = x

Therefore; f(x) - 1 = x - 1

f(x + 4) = x + 4

[tex]\frac{f(x) - 1}{f(x + 4)} = \frac{x-1}{x + 4}[/tex]

The piecewise function graph indicates that at x  → 1⁻, two points on the graph are; (0, 3), and (1, 4)

The slope of the graph therefore is; (4 - 3)/(1 - 0) = 1

The coordinates of the y-intercept is; (0, 3)

The function equation in slope-intercept form is therefore;

f(x) = x + 3

f(f(x) + 1) = x + 3 + 1 = x + 4

f(f(x) + 1) = x + 4

The limit value is as x tends to 1⁻, therefore;

The value of f(f(x) + 1) at x = 1 is; 1 + 4 = 5

Therefore;

The points of the function at f(6) are; (5, 2), and (7, 6)

The slope is; (6 - 2)/(7 - 5) = 2

The equation is; f(x) - 2 = 2·(x - 5) = 2·x - 10

f(x) - 2 = 2·x - 10

f(x) = 2·x - 10 + 2 = 2·x - 8

f(x) = 2·x - 8

f(6 + h) = 2·(6 + h) - 8 = 12 + 2·h - 8 = 4 + 2·h

f(6) = 2 × 6 - 8 = 4

f(6 + h) - f(6) = 4 + 2·h - 4 = 2·h

[tex]\frac{f(6 + h) - f(6)}{h} = \frac{2\cdot h}{h} = 2[/tex]

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