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schepotkina [342]
3 years ago
15

A recipe for 48 cookies use 2 and 2/3 cups of biscuit mix how many cookies are made from each cup of cookies

Mathematics
2 answers:
Nadya [2.5K]3 years ago
8 0

Answer:

B: 18

Step-by-step explanation:

48/2.66= 18

madreJ [45]3 years ago
5 0

Answer:

i believe the answer is B

Step-by-step explanation:

make 2 and 2/3 a decimal

2 divided by 3 will give you .6 repeating. just round to .8

then divide 48 by 2.8

it will give you 17.142

i just rounded it to 18... bc its so close to it

<u>hope this helps :)</u>

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3 0
3 years ago
An oak tree cast a shadow of 14 feet. A 30 foot flagpole casts a shadow of 60 how tall the Oak tree?
Alina [70]

Answer:

7 feet

Step-by-step explanation:

Oak tree = x

14 : x  as  60 : 30

Simplify the right ratio

14 : x  as 2:1

Get the same first number in both ratios by multiplying 2:1 by 7

14 : x  as 2(7) + 1(7)

14 : x as 14 : 7

Since the first number in each ratio is the same, we know that that x and 7 are equal

x = 7

Hope this helps, feel free to let me know if you would like a further explanation :)

5 0
3 years ago
Draw a figure thats less than 1/6
HACTEHA [7]
1 Whole is bigger than a fraction
1 part of the 6 equal groups is shaded

1/3
6 0
3 years ago
Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca
Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

\frac{dQ}{dt} = -rt

Which has the following solution:

Q(t) = Q(0)e^{-rt}

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that Q(6) = (1-0.03)Q(0) = 0.97Q(0). We use this to find r.

Q(t) = Q(0)e^{-rt}

0.97Q(0) = Q(0)e^{-6r}

e^{-6r} = 0.97

\ln{e^{-6r}} = \ln{0.97}

-6r = \ln{0.97}

r = -\frac{\ln{0.97}}{6}

r = 0.0051

So

Q(t) = Q(0)e^{-0.0051t}

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

Q(t) = Q(0)e^{-0.0051t}

0.5Q(0) = Q(0)e^{-0.0051t}

e^{-0.0051t} = 0.5

\ln{e^{-0.0051t}} = \ln{0.5}

-0.0051t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.0051}

t = 135.9

The half-life of the radioactive substance is 135.9 hours.

6 0
3 years ago
Find the product 20*9*15 tell which property you used
alukav5142 [94]
Okay so first what you would have to do is to read the question carefully and understand it because if you don't read the question, well the thing is you could get it wrong. and so first see what the question is and then understand it and once you'd understand it, solve it. So what you would have to do is tell the property is using and think about how the question is written and think of the properties you have studied. Well you might as well notice that its the associative property because if you don't have parenthesis around your problem, it wouldn't be as organized as it should be because if you don't follow the Associative Property, it could confuse you. And so the answer to this problem would be a more good or sensible answer because you used the properties to help you and so your final answer to this problem would be 2,700 as your final answer and keep in mind that if you use properties it would be a much easier problem to solve. And so thank for your question and have a blessed day and May God bless you and I hope this helped you out with your question you asked and so thank you again and so see you again. Bye !!!
7 0
3 years ago
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