Based on the amount that she paid in the first month, the amount Ronda will pay for the next month is<u> $396.</u>
When a loan is amortized, it means that one can pay it off by paying the same amount every period until they would have paid off both the loan and the associated interest.
The amortized amount contains:
- A portion going towards the principal(debt )
- A portion going towards the interest accumulated.
In conclusion, as the amount is the same every time, Ronda will have to pay the same amount of $396 the next month.
<em>Find out more at brainly.com/question/12256592. </em>
since all of the sides are not equal, it is not a regular polygon.
a regular polygon is when all of the sides are equal, and the same with the angles.
hope this helps you!! <3
Answer:
Step-by-step explanation:
Alright, lets get started.
The surface area of the figure given is : The area of all shapes together
The figure has 2 similar triangles and 3 similar rectangles.
The area of 1 triangle =

The area of 1 triangle = 
So, the area of both triangles = 
The area of 1 rectangle = 
The of 1 rectangle = 
So, the area of all 3 rectangles = 
So, total area of figure = 
hence the surface area is 536.6 square centimeters : Answer
Hope it will help :)
Answer:
![x= -\frac{ln [\frac{5P}{8000-P}]}{0.002}](https://tex.z-dn.net/?f=%20x%3D%20-%5Cfrac%7Bln%20%5B%5Cfrac%7B5P%7D%7B8000-P%7D%5D%7D%7B0.002%7D)
a) ![x= -\frac{ln [\frac{5*200}{8000-200}]}{0.002} =1027.062 \approx 1027](https://tex.z-dn.net/?f=%20x%3D%20-%5Cfrac%7Bln%20%5B%5Cfrac%7B5%2A200%7D%7B8000-200%7D%5D%7D%7B0.002%7D%20%3D1027.062%20%5Capprox%201027)
b) ![x= -\frac{ln [\frac{5*800}{8000-800}]}{0.002} =293.893 \approx 294](https://tex.z-dn.net/?f=%20x%3D%20-%5Cfrac%7Bln%20%5B%5Cfrac%7B5%2A800%7D%7B8000-800%7D%5D%7D%7B0.002%7D%20%3D293.893%20%5Capprox%20294)
Step-by-step explanation:
For this case we have the following function:

We can solve for x like this. First we can reorder the expression like this:



Now we can apply natura log on both sids and we got:
![ln[\frac{40000}{8000-P} -5] = ln e^{-0.002x}](https://tex.z-dn.net/?f=%20ln%5B%5Cfrac%7B40000%7D%7B8000-P%7D%20-5%5D%20%3D%20ln%20e%5E%7B-0.002x%7D)
![ln [\frac{5P}{8000-P}] = -0.002x](https://tex.z-dn.net/?f=%20ln%20%5B%5Cfrac%7B5P%7D%7B8000-P%7D%5D%20%3D%20-0.002x%20)
And if we solve for x we got:
![x= -\frac{ln [\frac{5P}{8000-P}]}{0.002}](https://tex.z-dn.net/?f=%20x%3D%20-%5Cfrac%7Bln%20%5B%5Cfrac%7B5P%7D%7B8000-P%7D%5D%7D%7B0.002%7D)
Part a
For this case we can replace P = 200 and see what we got for x like this:
![x= -\frac{ln [\frac{5*200}{8000-200}]}{0.002} =1027.062 \approx 1027](https://tex.z-dn.net/?f=%20x%3D%20-%5Cfrac%7Bln%20%5B%5Cfrac%7B5%2A200%7D%7B8000-200%7D%5D%7D%7B0.002%7D%20%3D1027.062%20%5Capprox%201027)
Part b
For this case we can replace P = 800 and see what we got for x like this:
![x= -\frac{ln [\frac{5*800}{8000-800}]}{0.002} =293.893 \approx 294](https://tex.z-dn.net/?f=%20x%3D%20-%5Cfrac%7Bln%20%5B%5Cfrac%7B5%2A800%7D%7B8000-800%7D%5D%7D%7B0.002%7D%20%3D293.893%20%5Capprox%20294)