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2 years ago

14

How can I use an equation with a single unknown to help understand with my billing

2 answers:

dedylja [7]2 years ago

4 0

Answer:

Almost any situation where there is an unknown quantity can be represented by a linear equation, like figuring out income over time, calculating mileage rates, or predicting profit. Many people use linear equations every day, even if they do the calculations in their heads without drawing a line graph.

Step-by-step explanation:

Medical billers use math to process bills and inform insurance companies about any pending charges for a patient's services. They need to find out the amount insurance companies will pay to cover a patient's bills. If the insurance claim is accepted, they make a payment to the healthcare provider.

geniusboy [140]2 years ago

3 0

Answer:

U just said a random thing

Step-by-step explanation:

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PLEASE HELP..................James is paid $18 an hour at his new job. He wishes to graph his total pay T, as a function of the

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450 AND 50.................

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3 years ago

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Luden [163]

Cone details:

height: h cm

radius: r cm

Sphere details:

radius: 10 cm

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

From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

<u>Using Pythagoras Theorem</u>

(a)

TO² + TU² = OU²

(h-10)² + r² = 10² [insert values]

r² = 10² - (h-10)² [change sides]

r² = 100 - (h² -20h + 100) [expand]

r² = 100 - h² + 20h -100 [simplify]

r² = 20h - h² [shown]

r = √20h - h² ["r" in terms of "h"]

(b)

volume of cone = 1/3 * π * r² * h

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

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (\sqrt{20h - h^2})^2 \ ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20h - h^2) (h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20 - h) (h) ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)$

To find maximum/minimum, we have to find first derivative.

(c)

<u>First derivative</u>

$\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )$

<u>apply chain rule</u>

$\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}$

<u>Equate the first derivative to zero, that is V'(x) = 0</u>

$\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0$

$\Longrightarrow \sf 40h-3h^2=0$

$\Longrightarrow \sf h(40-3h)=0$

$\Longrightarrow \sf h=0, \ 40-3h=0$

$\Longrightarrow \sf h=0,\:h=\dfrac{40}{3}$ <u />

<u>maximum volume:</u> <u>when h = 40/3</u>

$\sf \Longrightarrow max= \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )$

$\sf \Longrightarrow maximum= 1241.123 \ cm^3$

<u>minimum volume:</u> <u>when h = 0</u>

$\sf \Longrightarrow min= \dfrac{1}{3} \pi (0)^2(20-0)$

$\sf \Longrightarrow minimum=0 \ cm^3$

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