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nadya68 [22]
2 years ago
8

Mika bought 9 rolls of film to take 180 pictures on a field trip. Some rolls had 36 exposures and the rest had 12 exposures. How

many of each type did Mika buy
Mathematics
1 answer:
Juli2301 [7.4K]2 years ago
6 0

The number of each type Mika bought =x = 6.8 rolls; y= 2.2 rolls.

<h3>Calculation using ratio</h3>

The number of rolls of film Mika  bought =  9 rolls

The number of rolls that had 36 exposures = x

The number of rolls that had 12 exposures = y

The total number of exposures = 36 + 12 = 48

To find the exposure by one roll = 48/9 = 5.3

If one roll = 5.3 exposure

    x roll    = 36 exposure

Make x roll the subject of formula,

x roll = 36/5.3 = 6.8 rolls

  • If one roll = 5.3 exposure

    y roll    = 12 exposure

Make y roll the subject of formula,

y = 12/5.3 = 2.2 rolls

Therefore, the number of each type Mika bought = x = 6.8 rolls; y= 2.2 rolls.

Learn more about ratios here:

brainly.com/question/2328454

#SPJ1

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Move +2x to the other side. Sign changes from +2x to -2x.

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Answer:

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Step-by-step explanation:

Here, the given expression is:  (\frac{5}{6} a^{9}p^{5})  ^{3}

Now, starting from the outer most bracket.

As we know :

(abc)^{n}   = (a)^{n} \times (b)^{n}  \times (c)^{n}

and (a^m)^{n} = a^{(m \times n)}

⇒ (\frac{5}{6} a^{9}p^{5})  ^{3} = (\frac{5}{6})^{3} \times (a^{9})^{3}   \times (p^{5}) ^{3}\\

=\frac{125}{216}  \times (a)^{(9\times3) } \times (p)^{(5 \times 3)}\\= \frac{125}{216}  \times a^{(27) } \times p^{(15)}

Hence, the given expression  (\frac{5}{6} a^{9}p^{5})  ^{3} = \frac{125}{216}  \times a^{(27) } \times p^{(15)}

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