Answer:
I believe the answer is four there is twice as many horses so you would double the amount of original goats.
Answer:
The answer is "5051".
Step-by-step explanation:
Given equation:
![\bold{E(x)= 2(x+3)^2-3(x-1)(x+3)+(x-2)^2-31}\\\\](https://tex.z-dn.net/?f=%5Cbold%7BE%28x%29%3D%202%28x%2B3%29%5E2-3%28x-1%29%28x%2B3%29%2B%28x-2%29%5E2-31%7D%5C%5C%5C%5C)
put the 1, 2, 3, ........2019, 20202:
When E(1),
![E(1) =2(1+3)^2-3(1-1)(1+3)+(1-2)^2-31\\\\](https://tex.z-dn.net/?f=E%281%29%20%3D2%281%2B3%29%5E2-3%281-1%29%281%2B3%29%2B%281-2%29%5E2-31%5C%5C%5C%5C)
![=2(4)^2-3(0)(4)+(-1)^2-31\\\\=2(16)-0+1-31\\\\=32+1-31\\\\=2](https://tex.z-dn.net/?f=%3D2%284%29%5E2-3%280%29%284%29%2B%28-1%29%5E2-31%5C%5C%5C%5C%3D2%2816%29-0%2B1-31%5C%5C%5C%5C%3D32%2B1-31%5C%5C%5C%5C%3D2)
When E(2),
![E(2) =2(2+3)^2-3(2-1)(2+3)+(2-2)^2-31\\\\](https://tex.z-dn.net/?f=E%282%29%20%3D2%282%2B3%29%5E2-3%282-1%29%282%2B3%29%2B%282-2%29%5E2-31%5C%5C%5C%5C)
![=2(5)^2-3(1)(5)+(0)^2-31\\\\=2(25)-15+0-31\\\\=50-15-31\\\\=4](https://tex.z-dn.net/?f=%3D2%285%29%5E2-3%281%29%285%29%2B%280%29%5E2-31%5C%5C%5C%5C%3D2%2825%29-15%2B0-31%5C%5C%5C%5C%3D50-15-31%5C%5C%5C%5C%3D4)
When E(3),
![E(3) =2(3+3)^2-3(3-1)(3+3)+(3-2)^2-31\\](https://tex.z-dn.net/?f=E%283%29%20%3D2%283%2B3%29%5E2-3%283-1%29%283%2B3%29%2B%283-2%29%5E2-31%5C%5C)
![=2(6)^2-3(2)(6)+(1)^2-31\\\\=2(36)-36+1-31\\\\=72-66\\\\=6](https://tex.z-dn.net/?f=%3D2%286%29%5E2-3%282%29%286%29%2B%281%29%5E2-31%5C%5C%5C%5C%3D2%2836%29-36%2B1-31%5C%5C%5C%5C%3D72-66%5C%5C%5C%5C%3D6)
When E(2019),
![E(2019) =2(2019+3)^2-3(2019-1)(2019+3)+(2019-2)^2-31\\](https://tex.z-dn.net/?f=E%282019%29%20%3D2%282019%2B3%29%5E2-3%282019-1%29%282019%2B3%29%2B%282019-2%29%5E2-31%5C%5C)
![=2(2022)^2-3(2018)(2022)+(2017)^2-31\\\\=8,176,968-12,241,188+4,068,289-31 \\\\= 4038](https://tex.z-dn.net/?f=%3D2%282022%29%5E2-3%282018%29%282022%29%2B%282017%29%5E2-31%5C%5C%5C%5C%3D8%2C176%2C968-12%2C241%2C188%2B4%2C068%2C289-31%20%5C%5C%5C%5C%3D%204038)
When E(2020),
![E(2020) =2(2020+3)^2-3(2020-1)(2020+3)+(2020-2)^2-31\\\\](https://tex.z-dn.net/?f=E%282020%29%20%3D2%282020%2B3%29%5E2-3%282020-1%29%282020%2B3%29%2B%282020-2%29%5E2-31%5C%5C%5C%5C)
![=2(2023)^2-3(2019)(2023)+(2018)^2-31\\\\=8,185,058-12,253,311+4,072,324-31\\\\=4040\\](https://tex.z-dn.net/?f=%3D2%282023%29%5E2-3%282019%29%282023%29%2B%282018%29%5E2-31%5C%5C%5C%5C%3D8%2C185%2C058-12%2C253%2C311%2B4%2C072%2C324-31%5C%5C%5C%5C%3D4040%5C%5C)
![\ Calculate: \\\\ \bold{A= E(1)-E(2)+E(3)-E(4).....+E(2019)-E(2020)}\\\\A= 2-4+6-8.....+4038-4040](https://tex.z-dn.net/?f=%5C%20Calculate%3A%20%5C%5C%5C%5C%20%5Cbold%7BA%3D%20E%281%29-E%282%29%2BE%283%29-E%284%29.....%2BE%282019%29-E%282020%29%7D%5C%5C%5C%5CA%3D%202-4%2B6-8.....%2B4038-4040)
![A= (2+6....+4038)-(4+ 8+12...4040)\\](https://tex.z-dn.net/?f=A%3D%20%282%2B6....%2B4038%29-%284%2B%208%2B12...4040%29%5C%5C)
Formula:
![t_n= a+(n-1)d\\\\\ When \\ \\ T_n =4040 \\ a= 4 \\ d= 4\\4040= 4+(n-1)4\\4040=4+4n-4\\4n=4040\\n= \frac{4040}{4}\\n= 1010\\\\When \\\\T_n =4038 \\ a= 2 \\ d= 4\\4038= 2+(n-1)4\\4038=2+4n-4\\4038=-2+4n\\4036=4n \\ n= \frac{4036}{4}\\n= 1009\\\\](https://tex.z-dn.net/?f=t_n%3D%20a%2B%28n-1%29d%5C%5C%5C%5C%5C%20When%20%5C%5C%20%5C%5C%20%20T_n%20%3D4040%20%5C%5C%20a%3D%204%20%5C%5C%20d%3D%204%5C%5C4040%3D%204%2B%28n-1%294%5C%5C4040%3D4%2B4n-4%5C%5C4n%3D4040%5C%5Cn%3D%20%5Cfrac%7B4040%7D%7B4%7D%5C%5Cn%3D%201010%5C%5C%5C%5CWhen%20%5C%5C%5C%5CT_n%20%3D4038%20%5C%5C%20a%3D%202%20%5C%5C%20d%3D%204%5C%5C4038%3D%202%2B%28n-1%294%5C%5C4038%3D2%2B4n-4%5C%5C4038%3D-2%2B4n%5C%5C4036%3D4n%20%5C%5C%20n%3D%20%5Cfrac%7B4036%7D%7B4%7D%5C%5Cn%3D%201009%5C%5C%5C%5C)
Formula for sum:
![When, n= 1010\\ a=4 \\ l=4040\\ \\ S= \frac{n}{2}(2a+l)\\\\S= \frac{1010}{2} (2\times 4+4040)\\\\S= 505(8+4040)\\\\S_1= 2044240 \\\\\\](https://tex.z-dn.net/?f=When%2C%20n%3D%201010%5C%5C%20a%3D4%20%5C%5C%20l%3D4040%5C%5C%20%5C%5C%20S%3D%20%5Cfrac%7Bn%7D%7B2%7D%282a%2Bl%29%5C%5C%5C%5CS%3D%20%5Cfrac%7B1010%7D%7B2%7D%20%282%5Ctimes%204%2B4040%29%5C%5C%5C%5CS%3D%20505%288%2B4040%29%5C%5C%5C%5CS_1%3D%202044240%20%5C%5C%5C%5C%5C%5C)
![When, n= 1009\\ a=2 \\ l=4038\\ \\ S= \frac{n}{2}(2a+l)\\\\S= \frac{1009}{2} (2\times 2+4038)\\\\S= 504.5(4+4038)\\\\S_2= 2039189 \\\\\\](https://tex.z-dn.net/?f=When%2C%20n%3D%201009%5C%5C%20a%3D2%20%5C%5C%20l%3D4038%5C%5C%20%5C%5C%20S%3D%20%5Cfrac%7Bn%7D%7B2%7D%282a%2Bl%29%5C%5C%5C%5CS%3D%20%5Cfrac%7B1009%7D%7B2%7D%20%282%5Ctimes%202%2B4038%29%5C%5C%5C%5CS%3D%20504.5%284%2B4038%29%5C%5C%5C%5CS_2%3D%202039189%20%5C%5C%5C%5C%5C%5C)
![S= S_1-S_2\\\\S= 2044240 -2039189\\\\S= 5051](https://tex.z-dn.net/?f=S%3D%20S_1-S_2%5C%5C%5C%5CS%3D%202044240%20-2039189%5C%5C%5C%5CS%3D%205051)
The absolute value is = 5051
Review your notes and have a friend test you on them. Go over anything you had trouble on by reading it 10 times, writing it 10 times, and saying it out loud 2 times. It should help you remember and good luck! :)
Answer:
Simplified or Evaluated Expression:
-52u + 16
Step-by-step explanation:
Answer:
68º
Step-by-step explanation:
every quadrilateral has a total of 360º when adding all of the angles.
so since angle D and A are the same you add them along with angle B
110+110+72=292
then you subtract 360-292= 68º
Hope this helped!! Make sure to mark as brainliest :)