2t+5v=37 and 4t+3v=39
Solve the first for t
t=18.5-2.5v and use this t in the second equation...
4(18.5-2.5v)+3v=39
74-10v+3v=39
-7v=-35
v=$5.00 and since t=18.5-2.5v
t=$6.00
So tuna sandwiches cost $6.00 and vegetarian sandwiches cost $5.00.
Answer:
1)![a_{n}=a_{1}+(n-1)d\Rightarrow a_{n}=-1-2(n-1)\\a_{2}=-1-2(2-1)\Rightarrow a_{2}=-3\\a_{3}=-1-2(3-1)\Rightarrow a_{3}=-5\\(...)\\a_{10}=-1-2(10-1)=-19](https://tex.z-dn.net/?f=a_%7Bn%7D%3Da_%7B1%7D%2B%28n-1%29d%5CRightarrow%20a_%7Bn%7D%3D-1-2%28n-1%29%5C%5Ca_%7B2%7D%3D-1-2%282-1%29%5CRightarrow%20a_%7B2%7D%3D-3%5C%5Ca_%7B3%7D%3D-1-2%283-1%29%5CRightarrow%20a_%7B3%7D%3D-5%5C%5C%28...%29%5C%5Ca_%7B10%7D%3D-1-2%2810-1%29%3D-19)
2) ![a_{10}=4+8(10-1)\Rightarrow a_{10}=76](https://tex.z-dn.net/?f=a_%7B10%7D%3D4%2B8%2810-1%29%5CRightarrow%20a_%7B10%7D%3D76)
Step-by-step explanation:
1) To write an Arithmetic Sequence, as an Explicit Term, is to write a general formula to find any term for this sequence following this pattern:
![a_{n}=a_{1}+d(n-1)\Rightarrow \left\{\begin{matrix}a_{n}=n^{th}\: term\\a_1=1st \: term \\d=\: difference\\n=n^{th}\, term\end{matrix}\right.](https://tex.z-dn.net/?f=a_%7Bn%7D%3Da_%7B1%7D%2Bd%28n-1%29%5CRightarrow%20%5Cleft%5C%7B%5Cbegin%7Bmatrix%7Da_%7Bn%7D%3Dn%5E%7Bth%7D%5C%3A%20term%5C%5Ca_1%3D1st%20%5C%3A%20term%20%5C%5Cd%3D%5C%3A%20difference%5C%5Cn%3Dn%5E%7Bth%7D%5C%2C%20term%5Cend%7Bmatrix%7D%5Cright.)
<em>"Write an explicit formula for each explicit formula A(n)=-1+(n-1)(-2)"</em>
This isn't quite clear. So, assuming you meant
Write an explicit formula for each term of this sequence A(n)=-1+(n-1)(-2)
As this A(n)=-1+(n-1)(-2) is already an Explicit Formula, since it is given the first term
the common difference
let's find some terms of this Sequence through this Explicit Formula:
![a_{n}=a_{1}+(n-1)d\Rightarrow a_{n}=-1-2(n-1)\\a_{2}=-1-2(2-1)\Rightarrow a_{2}=-3\\a_{3}=-1-2(3-1)\Rightarrow a_{3}=-5\\(...)\\a_{10}=-1-2(10-1)=-19](https://tex.z-dn.net/?f=a_%7Bn%7D%3Da_%7B1%7D%2B%28n-1%29d%5CRightarrow%20a_%7Bn%7D%3D-1-2%28n-1%29%5C%5Ca_%7B2%7D%3D-1-2%282-1%29%5CRightarrow%20a_%7B2%7D%3D-3%5C%5Ca_%7B3%7D%3D-1-2%283-1%29%5CRightarrow%20a_%7B3%7D%3D-5%5C%5C%28...%29%5C%5Ca_%7B10%7D%3D-1-2%2810-1%29%3D-19)
2)
In this Arithmetic Sequence the common difference is 8, the first term value is 4.
Then, just plug in the first term and the common difference into the explicit formula:
![a_{10}=4+8(10-1)\Rightarrow a_{10}=76](https://tex.z-dn.net/?f=a_%7B10%7D%3D4%2B8%2810-1%29%5CRightarrow%20a_%7B10%7D%3D76)
The Number of boys in the class is: 18
Let's call "g" to the number of girls in the class.
The problem says that there is "w" fewer boys than girls, so we have:
g-w=18
Then, the number of girls in the class is:
g=18+w
Let's call "x" to the total number of studens in the class:
x=g+18
When we substitute g=18+w into x=g+18, we obtain:
x= (18+w)+18
The total number of studens in the class is:
x=w+36
Answer:
There are exactly 52.142857142857 weeks in the year 2019. This is equivalent to 52 weeks and 1 extra day, since there are 365 total days in 2019. Most years have 365 days, but a leap year has 366 days. That adds up to 52 weeks (where each week is exactly 7 days) PLUS 1 or 2 additional days.There are 52 complete weeks in a year. The year has 365 days, leaving one extra day. A leap year has 366 days, adding a second extra day. This makes 52 1/7 weeks in a normal year and 52 2/7 weeks in a leap year..
We can solve this pretty easily using a proportion where x equals the number of cloudy days in the year:
2/5 = x/365
Now cross multiply:
730 = 5x
And finally divide both aides by five:
146 = x
So 146 days of the year were cloudy.
Hope I helped, and let me know if you have any questions!