Mr. Ramirez can use many different group sizes, but the largest group size possible is 8. This results in 16/8=2 sixth-grade groups and 24/8=3 seventh-grade groups.
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
Let the numbers be n, n+2, n+4
Sum equals too= 13+2(n+4), which is 2n+21
a) Equation--> n+n+2+n+4= 2n+21
b) Solution--> 3n+6= 2n+21
=> n= 15
c) Second number--> 17 (15+2)
Third number--> 19 (15+4)
d) 15+15+2+15+4=30+21
=> 51= 51
So, the equation is true.
And pls mark me brainliesttt :)))
Answer:
B
Step-by-step explanation:
247 divided by 15 equals 16 and a whole bunch of nunbers
