Let
x-------> the number of dinner
y-------> the number of lunch
we know that
-------> equation A
------> equation B
Substitute equation B in equation A
![8[y]+5y \leq 42](https://tex.z-dn.net/?f=8%5By%5D%2B5y%20%5Cleq%2042)
![13y \leq 42](https://tex.z-dn.net/?f=13y%20%5Cleq%2042)
![y \leq 42/13](https://tex.z-dn.net/?f=y%20%5Cleq%2042%2F13)
![y \leq 3.23](https://tex.z-dn.net/?f=y%20%5Cleq%203.23)
so
the greatest number of lunch is ![y=3](https://tex.z-dn.net/?f=y%3D3)
![x=y](https://tex.z-dn.net/?f=x%3Dy)
Hence
the greatest number of dinner is ![x=3](https://tex.z-dn.net/?f=x%3D3)
therefore
the greatest number of meals is
![x+y=3+3=6](https://tex.z-dn.net/?f=x%2By%3D3%2B3%3D6)
<u>the answer is</u>
![6](https://tex.z-dn.net/?f=6)
You can use a graphing paper or a software to create a scatter plot. I used MS Excel to plot the points. You simply have to type the values and create a graph. The x-axis is the number of years and the y-axis is the savings account balance.
Answer:
0,07 0,7 0,704 0,74 0,744
Step-by-step explanation:
When u make it 100 times bigger, its clear.
7 70 704 740 744
Answer:13
Step-by-step explanation:
3+9+1=13