MATHEMATICS HIGH SCHOOL

Each day that a library book is kept past its due date, a $0.30 fee is charged at midnight. Which ordered pair is a viable solution if x represents the number of days that a library book is late and y represents the total fee?Answers:
(–3, –0.90)
(–2.5, –0.75)
(4.5, 1.35)
(8, 2.40)

Answers

Answer 1
Answer: The answer is (8, 2.40).

We can deduce this because if x is negative, it means the book is not late; it's early! We can cross out the first two answers. We know the third one is wrong because the fee is charged on midnight, so the fee for 4.5 days late would be the fee for 4 days late, which would be 1.20, not 1.35. for 8 days, the cost would be (8*0.30) = 2.40, which is what we have for y in the pair. Therefore, (8, 2.40) is the answer.

Hope this helps! :)
Answer 2
Answer:

Answer:

The answer is (8, 2.40).

Step-by-step explanation:

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What is the unit rate for 840 feet in 40 seconds?

Answers

840 feet / 40 seconds = 21 feet per second.

Write sentence as inequality. Plz help
A number x is less than 5 and greater than 3.

Answers

Answer:

3<x<5

Step-by-step explanation:

Order the following masses from least to greatest: 500 g, 50 kg, 5,000 mg

Answers

The order is
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Answer:

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I need help with 1+1 First answer gets Brainliest(Please explain how you got the answer)

Answers

Answer:

2

Step-by-step explanation:

HELP. I don't know if I'm supposed to put in theorems or?? It's a fill in the blank for a,b,c,d, and e.
Pretty sure A and B are "Given" but not 100%.

Answers

I hope this helps you

What is the y-coordinate of the solution? Round to the nearest hundredth.
( 4x-y= 2
2x+9y= -12)

Answers

Hi \n \n \begin{cases}4x-y=2~~----\ \textgreater \ (I) \n 2x+9y=- 12~--\ \textgreater \ (II)\end{cases} \n \n \underline{Resolution ~} \n \n~~~~~~4x-y=2 \n =\ \textgreater \ \boxed{y=4x-2}~~--\ \textgreater \ \textcircled{1} \n \n \mathbb{tttttttttttttttttttttttttttttttttttt} \n\textcircled{1}~in (II) \n \n2x+9\underbrace{y}_(4x-2)=-12 \n \n 2x+9(4x-2)=-12 \n \n 2x+36x-18=-12 \n \n \boxed{\boxed{x=0,16}} \n \n \mathbb{tttttttttttttttttttttttttttttttttttt} \n =\ \textgreater \ y=4x-2 \n ~~~~~~y=4(0,16)-2 \n ~~~~~~\boxed{\boxed{y=-1,36}}\n \n \n .

\therefore~(x;y)=(0,16; -1,36)
y=-1.4
all you had to do was the elimination method
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