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jeka94
3 years ago
10

Tom has 28 milk chocolates and Emily has 24 dark chocolates. They have to divide these chocolates into small packets that each h

ave the same
number of dark chocolates and the same number of milk chocolates. They also want to use all of the chocolates,
The maximum number of packets they can make is
To make the maximum number of packets, each packet should have
milk chocolates and
dark chocolates
Mathematics
2 answers:
Aliun [14]3 years ago
8 0

Answer:

The answer is 4.

Step-by-step explanation:

RSB [31]3 years ago
6 0

Answer:

4 milk chocolatero need to be removed to have even amounts

24+24= 48

24 packets with 1 of each(2 count) is one possibility.

You can have less packets with more chocolates inside like 12 packets with 4 count(2 of each)

Any combination with result 48. To use al chocolatesl : 4 packets:Each packet has 6dark and 7milk choc

Step-by-step explanation:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Helppp please show work
Mamont248 [21]

Step-by-step explanation:

<u>Given</u><u> </u><u>that</u><u>:</u><u>-</u>

  • Side AC measures 6 units
  • Side BC measures 9.5 units
  • Side AB measures 11 units

<u>To</u><u> </u><u>find</u><u>:</u><u>-</u>

  • The cosB as a decimal rounded to the nearest hundredth

<u>We</u><u> </u><u>know</u><u> </u><u>that</u><u>, </u>

{ \bigstar \: { \underline{ \boxed{ \cos( \theta ) \:  =  \sf\frac{adjecent \: side}{hypotenuse}  }}}}

<u>Solution</u><u>:</u>

\longrightarrow \rm  \cos(B)  =  \dfrac{BC}{AB}

\longrightarrow \rm  \cos(B)  =  \dfrac{9.5}{11}

\longrightarrow \rm  \cos(B)  =  \dfrac{95}{110}

\longrightarrow \rm  \cos(B)  =  0.8 63

\longrightarrow  \rm  \cos(B)   \approx \: 0.9

<u>Therefore:</u>

  • Cos B equals 0.9

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