Endometriosis is an inflammatory disease caused by endometrial cells (tissue that lines the uterus) that, instead of being expelled during menstruation, move in the opposite direction and fall into the ovaries or abdominal cavity, where they multiply and to bleed.

<h3>What is an ovarian cyst?</h3>

An ovarian cyst is a fluid-filled sac that forms in or around the ovary, which can cause symptoms such as pain in the pelvic region, irregular menstruation, or bleeding outside the menstrual period.

With this information, we can conclude that she would likely be diagnosed with endometriosis or an ovarian cyst.

Learn more about endometriosis in brainly.com/question/14596231

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8 0

1 year ago

What are two measures of a physical health

juin [17]

Hello! The answer to your question would be as followed:

Two measures of physical health could be <u><em>a fitness-gram testing and seeing how long you last through it</em></u>

6 0

2 years ago

Differentiate the following functions (i) x(1+x)^3

statuscvo [17]

Answer:

$\displaystyle y' = (1 + x)^2(4x + 1)$

General Formulas and Concepts:

<u>Algebra I</u>

Terms/Coefficients
Functions
Function Notation
Factoring

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Explanation:

<u>Step 1: Define</u>

<em>Identify</em>

y = x(1 + x)³

<u>Step 2: Differentiate</u>

Product Rule [Derivative Rule - Chain Rule]: $\displaystyle y' = \frac{d}{dx}[x] \cdot (1 + x)^3 + x \cdot \frac{d}{dx}[(1 + x)^3] \cdot \frac{d}{dx}[1 + x]$
Derivative Property [Addition/Subtraction]: $\displaystyle y' = \frac{d}{dx}[x] \cdot (1 + x)^3 + x \cdot \frac{d}{dx}[(1 + x)^3] \cdot (\frac{d}{dx}[1] + \frac{d}{dx}[x])$
Basic Power Rule: $\displaystyle y' = x^{1 - 1} \cdot (1 + x)^3 + x \cdot 3(1 + x)^{3 - 1} \cdot (0 + x^{1 - 1})$
Simplify: $\displaystyle y' = (1 + x)^3 + 3x(1 + x)^2$
Factor: $\displaystyle y' = (1 + x)^2 \bigg[ (1 + x) + 3x \bigg]$
Combine like terms: $\displaystyle y' = (1 + x)^2(4x + 1)$