Answer:-5x-8y=-5
Step-by-step explanation:
add right sides of equation and even them with result of adding right sides of equation
4x-4y-9x-4y=-2-3
-5x-8y=-5
First, convert all of the cm measurements to m measurements (so they are all the same unit measurement)
2000 cm = 20 m 800 cm = 8m
<u>Total Perimeter </u>(Note that circumference of a semi-circle is 2 π r/2 = π r)
Add up the lengths of all of the outside edges. I am going to start on the top and move counter-clockwise:
40 + π (10) + 8 + 25 + 8 + (40 - 25 - 10) + 8 + 10 + 8 + π(10)
= 40 + 10π + 41 + (5) + 26 + 10π
= 112 + 20π
= 112 + 62.8
= 174.8
Answer: 174.8 m
<u>Total Area</u>
Split the picture into 5 sections (2 semi-circles, top rectangle, bottom left rectangle, and bottom right rectangle). Find the area for each of those sections and then add their areas together to find the total area.
2 semi-circles is 1 Circle: A = π · r² ⇒ A = π(20/2)² = π(10)² = 100π ≈ 314
top rectangle: A = L x w ⇒ A = 40 x 20 = 800
bottom left rectangle: A = L x w ⇒ A = 25 x 8 = 200
bottom right rectangle: A = L x w ⇒ A = 10 x 8 = 80
Total = 314 + 800 + 200 + 80 = 1394
Answer: 1394 m²
Well... how about pay the cost and enjoy the pizza and drinks.
Assume the second number is x, then the other two are x - 2, x + 2
So x-2 + x + x + 2 = 51
-> 3x = 51
-> x = 17
The question is incomplete. The complete question is :
The population of a certain town was 10,000 in 1990. The rate of change of a population, measured in hundreds of people per year, is modeled by P prime of t equals two-hundred times e to the 0.02t power, where t is measured in years since 1990. Discuss the meaning of the integral from zero to twenty of P prime of t, d t. Calculate the change in population between 1995 and 2000. Do we have enough information to calculate the population in 2020? If so, what is the population in 2020?
Solution :
According to the question,
The rate of change of population is given as :
in 1990.
Now integrating,

![$=\frac{200}{0.02}\left[e^{0.02(20)}-1\right]$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B200%7D%7B0.02%7D%5Cleft%5Be%5E%7B0.02%2820%29%7D-1%5Cright%5D%24)
![$=10,000[e^{0.4}-1]$](https://tex.z-dn.net/?f=%24%3D10%2C000%5Be%5E%7B0.4%7D-1%5D%24)
![$=10,000[0.49]$](https://tex.z-dn.net/?f=%24%3D10%2C000%5B0.49%5D%24)
=4900





This is initial population.
k is change in population.
So in 1995,



In 2000,


Therefore, the change in the population between 1995 and 2000 = 1,163.