Ax=aw-z-y
divide by x
a=awx-z-y
a=awx-zy
By using two arbitrary points on the line, we can find the slope using the slope formula:
We'll use the points (0, 4) and (5, 1) for this.
1 - 4 / 5 - 0
-3 / 5
The slope of the line is -3/5.
D, it increases then goes at a constant speed
Answer:
The number that comes after 144five is:
= 200five.
Step-by-step explanation:
Adding 1 to 144 base 5 will result in:
144
+ 1
= 200
b) To obtain the next number that comes after 144five, add 1five to 144five. Since the numbers are in base 5, 1five added to 4five will result in 0 with 1 carried backward. When 1 is added to the next 4, the result will be 0 with 1 carried backward. 1 added to 1 = 2, all in base 5. Figures in base 5 cannot exceed 4. The usual numbers for a base 5 operation are 0, 1, 2, 3, and 4.
As the question states, let r be the number of hours worked at the restaurant, and y be the number of hours of yard work.
We know that she can only work a maximum of 15 hours per work total, and that at she must work at least 5 hours in the restaurant.
Therefore:
r + y ≤ 15
r ≥ 5
We also know that she wants to earn at least 120 dollars, earning $8/hr in the restaurant and $12/hr in the yard:
8r + 12y ≥ 120
What is the maximum of hours Lia can work in the restaurant and still make at leas 120 hours?
Lia's parents won't let her work more than 15 hours, so we know that the answer won't be higher than 15.
If she worked all 15 hours in the restaurant, she would make 8*15 = 120.
The maximum number of hours she can work in the restaurant is therefore 15 hours
What is the maximum amount of money Lia can earn in a week?
Lia has to work a minimum of 5 hours in the restaurant. She makes more money doing yard work, so she should devote the rest of her available work hours to yard work.
That means that, given her 15 hour work limit, she will maximize her income by working 5 hours in the restaurant and 10 hours in the yard.
5*8 + 10*12 = 40 + 120 = 160
The most she can make is 160 dollars, working 5 hours in the restaurant and 10 hours in the yard
