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Question
- if he uses 0.8 gb per day, how many days can he use the data before reaching the limit?
- how much data will he have left after 12 days at the rate of 0.6 gb per day?
1v) a student works two jobs. job a pays $12 per hour; job b pays $15 per hour. she wants to earn at least $450 this week but can work no more than 35 hours total.
- write a system of inequalities to represent the total hours and total earnings constraints.
- if she works 20 hours at job a, how many hours must she work at job b to reach her earning goal?
- what are all possible combinations of hours at job a and job b that meet both constraints?
v) a chemistry lab requires that a chemical solution be kept at a temperature between 15°c and 25°c for safe use. temperatures below 15°c or above 25°c can cause dangerous reactions.
- represent the safe temperature range as an interval.
- if the temperature is currently 12°c, how much must it increase to reach the safe range?
- if the temperature is 28°c, is it within the safe range? if not, how much must it decrease?
IV
1.
Step1: Define variables
Let $x$ be the number of hours worked at Job A and $y$ be the number of hours worked at Job B.
Step2: Write total - hours constraint
The total number of hours worked is $x + y$, and it cannot exceed 35 hours. So, $x + y\leq35$.
Step3: Write total - earnings constraint
Job A pays $12 per hour and Job B pays $15 per hour, and the student wants to earn at least $450. So, $12x + 15y\geq450$.
The system of inequalities is
, where $x\geq0$ and $y\geq0$ (since the number of hours cannot be negative).
Step1: First, find earnings from Job A
If she works 20 hours at Job A, her earnings from Job A are $12\times20 = 240$ dollars.
Step2: Then, find remaining earnings
She wants to earn at least $450$, so the remaining amount she needs to earn is $450 - 240=210$ dollars.
Step3: Calculate hours at Job B
Since Job B pays $15 per hour, the number of hours $y$ she needs to work at Job B is $\frac{210}{15}=14$ hours.
Step1: Rewrite inequalities
From $x + y\leq35$, we have $y\leq - x + 35$. From $12x+15y\geq450$, we can rewrite it as $y\geq-\frac{4}{5}x + 30$. Also, $x\geq0$ and $y\geq0$.
Step2: Find intersection points
Intersection of $y=-x + 35$ and $y =-\frac{4}{5}x+30$:
$-x + 35=-\frac{4}{5}x + 30$
$-x+\frac{4}{5}x=30 - 35$
$-\frac{1}{5}x=-5$, so $x = 25$ and $y=10$.
Intersection of $y=-x + 35$ and $x = 0$ gives $(0,35)$. Intersection of $y=-x + 35$ and $y = 0$ gives $(35,0)$. Intersection of $y=-\frac{4}{5}x + 30$ and $x = 0$ gives $(0,30)$. Intersection of $y=-\frac{4}{5}x + 30$ and $y = 0$ gives $(37.5,0)$ (but considering the non - negativity and the total - hours constraint, we ignore $x>35$).
The possible combinations are non - negative integer pairs $(x,y)$ such that $-\frac{4}{5}x + 30\leq y\leq - x + 35$, $x\geq0$ and $y\geq0$.
We can list the pairs by considering integer values of $x$ from 0 to 35:
For $x = 0$, $30\leq y\leq35$ (pairs $(0,30),(0,31),(0,32),(0,33),(0,34),(0,35)$);
For $x = 5$, $26\leq y\leq30$ (pairs $(5,26),(5,27),(5,28),(5,29),(5,30)$);
For $x = 10$, $22\leq y\leq25$ (pairs $(10,22),(10,23),(10,24),(10,25)$);
For $x = 15$, $18\leq y\leq20$ (pairs $(15,18),(15,19),(15,20)$);
For $x = 20$, $14\leq y\leq15$ (pairs $(20,14),(20,15)$);
For $x = 25$, $y = 10$ (pair $(25,10)$);
For $x$ from 26 to 35, $y$ values are non - existent or non - valid according to the inequalities.
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