Answer:

Rounded to the nearest hundredths: 11.69.
Step-by-step explanation:
I would use the Pythagorean theorem for this problem.
The difference between the highest point and the lowest point of AD is 9.8-7.2 = 2.6, so that would be the height of the triangle. The length/base of the triangle would be 11.4.
Now, just solve using Pythagorean's theorem:

Rounded to the nearest hundredths: 11.69.
I hope this helped you.
The classifications of the functions are
- A vertical stretch --- p(x) = 4f(x)
- A vertical compression --- g(x) = 0.65f(x)
- A horizontal stretch --- k(x) = f(0.5x)
- A horizontal compression --- h(x) = f(14x)
<h3>How to classify each function accordingly?</h3>
The categories of the functions are given as
- A vertical stretch
- A vertical compression
- A horizontal stretch
- A horizontal compression
The general rules of the above definitions are:
- A vertical stretch --- g(x) = a f(x) if |a| > 1
- A vertical compression --- g(x) = a f(x) if 0 < |a| < 1
- A horizontal stretch --- g(x) = f(bx) if 0 < |b| < 1
- A horizontal compression --- g(x) = f(bx) if |b| > 1
Using the above rules and highlights, we have the classifications of the functions to be
- A vertical stretch --- p(x) = 4f(x)
- A vertical compression --- g(x) = 0.65f(x)
- A horizontal stretch --- k(x) = f(0.5x)
- A horizontal compression --- h(x) = f(14x)
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I think it’s B) V27 i could me wrong
The rearranged formula for volume in terms of radius,r is
.
What is volume?
- Volume is a measure of enthralled three-dimensional space.
- It's frequently quantified numerically using SI-deduced units or by colorful Homeric units( similar to the gallon, quart, boxy inch).
- The description of length( cubed) is interrelated with volume. The volume of a vessel is generally understood to be the capacity of the vessel; i.e., the quantum of fluid( gas or liquid) that the vessel could hold, rather than the quantum of space the vessel itself displaces.
- In ancient times, volume is measured using analogous-structured natural holders and latterly on, standardized holders.
- Some simple three-dimensional shapes can have their volume fluently calculated using computation formulas.
- Volumes of more complicated shapes can be calculated with integral math if a formula exists for the shape's boundary.
Volume, V = 
Rearranging,
r =![\sqrt[3]{\frac{V}{\pi } }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%5Cfrac%7BV%7D%7B%5Cpi%20%7D%20%7D)
The rearranged formula for volume in terms of radius,r is
.
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