Answer:
![x = 8](https://tex.z-dn.net/?f=x%20%3D%208)
x = 8 when y = 8
Step-by-step explanation:
explanation is in the image
X=4,Y=7
Step-by-step explanation:
You have to do pemdas method and combine all the like terms together
Answer:
3i + 10
Step-by-step explanation:
4i + 3 - i + 7
= (4i - i) + (3 + 7)
= 3i + 10
Answer:
(a) $4250,
(b) $5225.
Step-by-step explanation:
Let n represent number of months.
We have been given that John saves $3500 the first month and every month later decreases it by $75.
First of all, we will find formula of John's savings in n months using arithmetic progression, where, 1st term is 3500 and common difference is 75.
, where,
n = Number of terms in a sequence,
d = Common difference.
![a_n=3500+(n-1)75](https://tex.z-dn.net/?f=a_n%3D3500%2B%28n-1%2975)
![a_n=3500+75n-75](https://tex.z-dn.net/?f=a_n%3D3500%2B75n-75)
![a_n=3425+75n](https://tex.z-dn.net/?f=a_n%3D3425%2B75n)
(a) To find John's savings in 11 months, we will substitute
in above formula.
![a_{11}=3425+75(11)](https://tex.z-dn.net/?f=a_%7B11%7D%3D3425%2B75%2811%29)
![a_{11}=3425+825](https://tex.z-dn.net/?f=a_%7B11%7D%3D3425%2B825)
![a_{11}=4250](https://tex.z-dn.net/?f=a_%7B11%7D%3D4250)
Therefore, John would have saved $4250 in the 11th month.
(b) To find John's savings after 2 years, we will substitute
in above formula.
![a_{24}=3425+75(24)](https://tex.z-dn.net/?f=a_%7B24%7D%3D3425%2B75%2824%29)
![a_{24}=3425+1800](https://tex.z-dn.net/?f=a_%7B24%7D%3D3425%2B1800)
![a_{24}=5225](https://tex.z-dn.net/?f=a_%7B24%7D%3D5225)
Therefore, John would have saved $5225 in 2 years.