The section of the CPT coding manual is the code 40490 is listed in is "digestive system."
<h3>What is CPT codes?</h3>
CPT is the language used by providers and payers when billing medical procedures and services for reimbursement.
Some key features regarding the CPT codes are-
- CPT or Current Procedural Terminology, is a set of medical codes used to describe the procedures and services performed by physicians, allied health care professionals, provided by trained practitioners, healthcare facilities, outpatient facilities, and laboratories.
- CPT codes, in particular, are used to notify services and procedures to both federal and private payers for repayment of rendered healthcare services.
- CPT codes were developed by the American Medical Association (AMA) in 1966 to standardize reporting of medical, surgery, and diagnosis and treatment procedures and services performed in both inpatient and outpatient settings.
- Each CPT system defines a written description of a process or service, removing the need for the patient's subjective interpretation of what was provided.
To know more about the CPT codes, here
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Answer:
x = 4.
Step-by-step explanation:
log(x + 6)- log(x - 3) = 1
Using the law log a - log b = log a/b:
log (x + 6)/ (x - 3) = 1
Removing logs:
(x + 6)/ (x - 3) = 10
x+ 6 = 10x - 30
9x = 36
x = 4 (answer).
Hello macelynn190!
To solve this question we need to substitute the given values of 'a' & 'b' in the equation & then simplify it.
Given,
Now, substitute this in the place of 'a' & 'b'..
Lastly, simplify it...
- The answer is <u>-</u><u>9</u><u>/</u><u>2</u><u> </u><u>or </u><u>-</u><u>4</u><u>.</u><u>5</u>
__________________
Hope it'll help you!
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Answer:
look at the photo..................
Each of these roots can be expressed as a binomial:
(x+1)=0, which solves to -1
(x-3)=0, which solves to 3
(x-3i)=0 which solves to 3i
(x+3i)=0, which solves to -3i
There are four roots, so our final equation will have x^4 as the least degree
Multiply them together. I'll multiply the i binomials first:
(x-3i)(x+3i) = x²+3ix-3ix-9i²
x²-9i²
x²+9 [since i²=-1]
Now I'll multiply the first two binomials together:
(x+1)(x-3) = x²-3x+x-3
x²-2x-3
Lastly, we'll multiply the two derived terms together:
(x²+9)(x²-2x-3) [from the binomial, I'll distribute the first term, then the second term, and I'll stack them so we can simply add like terms together]
x^4 -2x³-3x²
<u> +9x²-18x-27</u>
x^4-2x³+6x²-18x-27
