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Ostrovityanka [42]
3 years ago
12

Ally records the time she spends doing homework. She rounds each time to the nearest ten minutes. If Ally spends 37 minutes on h

er homework, how does she
record this time?
Please answer!!!
Mathematics
2 answers:
fredd [130]3 years ago
8 0
She will most likely round to 40 since 37 is closet to that
Rama09 [41]3 years ago
5 0

Answer:

40 minutes?

Step-by-step explanation:

If she was rounding to the nearest 10? I am not sure .

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Courier charges for packages to a certain destination are 65 cents for the first 250 grams and 10 cents for each additional 100
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Well, drop 65 cents (the first 250 grams) to get 90 remaining cents. These are 900 additional grams. So, the mass is 0.25+0.9=1.15 kg.
4 0
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Wholly didn't know how to simplify 8/16
NemiM [27]
8/16=1/2
You divide the numerator and denominator by 8
I hope this helps!
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Im toooo bad at math plz help
balandron [24]

Answer:

-3/22

Step-by-step explanation:

Mulitiply straight across.  Num * num/ denom * denom.  

Now you have -6/44.  Divide both by 2 to simplify into simipliest form.  

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2 years ago
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Answer:

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Step-by-step explanation:

In pic

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3 years ago
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Find the exponential function that passes through the points (2,80) and (5,5120)​
juin [17]

Answer:

  y = 5·4^x

Step-by-step explanation:

If you have two points, (x1, y1) and (x2, y2), whose relationship can be described by the exponential function ...

  y = a·b^x

you can find the values of 'a' and 'b' as follows.

Substitute the given points:

  y1 = a·b^(x1)

  y2 = a·b^(x1)

Divide the second equation by the first:

  y2/y1 = ((ab^(x2))/(ab^(x1)) = b^(x2 -x1)

Take the inverse power (root):

  (y2/y1)^(1/(x2 -x1) = b

Use this value of 'b' to find 'a'. Here, we have solved the first equation for 'a'.

  a = y1/(b^(x1))

In summary:

  • b = (y2/y1)^(1/(x2 -x1))
  • a = y1·b^(-x1)

__

For the problem at hand, (x1, y1) = (2, 80) and (x2, y2) = (5, 5120).

  b = (5120/80)^(1/(5-2)) = ∛64 = 4

  a = 80·4^(-2) = 80/16 = 5

The exponential function is ...

  y = 5·4^x

3 0
3 years ago
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