Question

A plan that costs $29.99 another company $19.99 and $0.35 during nights and weekends for what numbers of night and weekend does the second company's plan cost more then the first company's plan

Answer 1

Answer:

29 minutes more

Step-by-step explanation:

Let m represent minutes

changing the statement to algebra, since the second company charges a different rate at night and weekend  we have the equation below;

$19.99 + $0.35m > $29.99

Subtract 19.99 from both sides to isolate m and we have;

$19.99 -$19.99 + $0.35 > $29.99 - $19.99

= $0,35m > $10.00

Divide both side by 0.35 to obtain the value of m;

$\frac{0.35m}{0.35}$ > $\frac{10}{0.35}$

= m > 28.57

m ⩾ 29 minutes

The second company's will be twenty nine minutes or more costlier than the first company

Answer 2

Answer:

$\ 0.35n>\ 29.99-\ 19.99\\\ 0.35n>\ 10\\n>28.57$

Approximately:

n≥29

So the number of night and weekend that the second company's plan cost more then the first company's plan are ≥ 29

Step-by-step explanation:

let say n is the numbers of night and weekend that  the second company's plan cost more then the first company's plan.

Since second company has $19.99 and $0.35 during nights and weekend and it is grater than the first company which costs $29.99.

According to the above condition, We will get the equation:

$\ 19.99+\ 0.35n>\ 29.99$

Solving the above equation:

$\ 0.35n>\ 29.99-\ 19.99\\\ 0.35n>\ 10\\n>28.57$

Approximately:

n≥29

So the number of night and weekend that the second company's plan cost more then the first company's plan are ≥ 29