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Alina [70]
3 years ago
7

-3 ¼ divided by -1/8

Mathematics
2 answers:
hram777 [196]3 years ago
6 0

\purple{ \tt{ \huge{  \: ✨Answer ✨ \: }}}

\:  \:  \:  \:  \:  \:  \:  \:  \:  \orange{ \boxed{ \boxed{ \huge{ \tt{{ \: 26 \: }}}}}}

Marat540 [252]3 years ago
4 0

Answer:

The answer is 26

Step-by-step explanation:

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salma purchased a prepaid phone card for $20. long distance calls cost 24 cents a minute using this card. salma used her card on
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We can set up an equation to solve this problem, but first we need to write out what we know.

$20 overall
$0.24 every minute
$13.52 remaining on the card

Now that we know our information, we can set it up in an equation. 

20 - 0.24x = 13.52

The 20 represents $20 overall when she first got the phone card.
We are then subtracting $20 from how must it costs a minute (which is 24 cents). The 'x' indicates the number we are trying to find (how many minutes her call lasted). Lastly, 13.52 is the result of everything, since she has $13.52 remaining on the card.

We can now solve the equation:

20 - 0.24x = 13.52
-0.24x = 13.52 - 20 /// subtract 20 from each side
-0.24x = -6.48 /// simplify
x = 27 /// divide each side by -0.24

Our solution is: x = 27.

-----

An easier way to solve this problem would be to first, subtract the total amount of money she had on the card when she first got it, and then the remaining total she ended up with.

$20 - $13.52 = $6.48

So, she spent a total of $6.48 on long distance calls, but since we are looking for how many minutes, we need to divide the total she spent and how much it costs per minute.

6.48 ÷ 24 = 27

We receive the same amount of minutes spent just like we did the last way we solved.

-----

Salma spent 27 minutes on the phone. 
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b) The remaining fence, after the two sides of length x are fenced, is 200-2x. That is the length of the side parallel to the building. The product of the lengths parallel and perpendicular to the building is the area of the playground:
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a) A(50) = 50(200 -2·50) = 50·100 = 5000 . . . . m²
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c) The equation makes no sense if either length (x or 200-2x) is negative, so a reasonable domain is (0, 100). For x=0 or x=100, the playground area is zero, so we're not concerned with those cases, either. Those endpoints could be included in the domain if you like.
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