$25.75 is how much money she saved up each week
Solution:
Given that the point P lies 1/3 along the segment RS as shown below:
To find the y coordinate of the point P, since the point P lies on 1/3 along the segment RS, we have
Using the section formula expressed as
![[\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bmx_2%2Bnx_1%7D%7Bm%2Bn%7D%2C%5Cfrac%7Bmy_2%2Bny_1%7D%7Bm%2Bn%7D%5D)
In this case,
where
Thus, by substitution, we have
![\begin{gathered} [\frac{1(2)+2(-7)}{1+2},\frac{1(4)+2(-2)}{1+2}] \\ \Rightarrow[\frac{2-14}{3},\frac{4-4}{3}] \\ =[-4,\text{ 0\rbrack} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5B%5Cfrac%7B1%282%29%2B2%28-7%29%7D%7B1%2B2%7D%2C%5Cfrac%7B1%284%29%2B2%28-2%29%7D%7B1%2B2%7D%5D%20%5C%5C%20%5CRightarrow%5B%5Cfrac%7B2-14%7D%7B3%7D%2C%5Cfrac%7B4-4%7D%7B3%7D%5D%20%5C%5C%20%3D%5B-4%2C%5Ctext%7B%200%5Crbrack%7D%20%5Cend%7Bgathered%7D)
Hence, the y-coordinate of the point P is
So we can start with the full of possibilities and eliminate them one by one.
The full set is {0,1,2,3,4,5,6,7,8,9}.
Now we know that any prime greater than 2 is odd as otherwise it would have 2 as a factor, so we can eliminate all of these digits that would be an even number, leaving:
{1,3,5,7,9}
We also know that any prime greater than 5 cannot be a multiple of 5 and that all numbers with 5 in the digits are a multiple of 5, so we can eliminate 5.
{1,3,7,9}
We know that 11,13,17 and 19 are all primes, so we cannot eliminate any more of these, leaving the set:
{1,3,7,9} as our answer.
Graph 1: Domain = 2 <= x < 12
Graph 1: Range = {2,4}
Graph 2: Domain = 3 <= x <= 10
Graph 2: Range = -6 <= y <= 4
Graph 3: Domain = 4 < x < 11
Graph 4: Range = -6 <= y < 2
Answer:
I think the answer is 2n-5
(-3n + 2)+(5n - 7)
-3n+5n=2n
2+(-7)=-5
