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ankoles [38]
3 years ago
15

Solve each calculation. Be sure to report your answer with the correct number of significant figures.

Mathematics
2 answers:
Serjik [45]3 years ago
5 0

Question:

Solve each calculation. Be sure to report your answer with the correct number of significant figures.

5.61000 dg  × 1.1010 dg  

12.0 m  ÷ 3.1415 m

Answer:

1. 5.61000dg  * 1.1010dg = 6.177 dg^2

2. \frac{12}{3.1415} = 3.8198

Step-by-step explanation:

1. 5.61000 dg  × 1.1010 dg  

We start by representing each digit using scientific notation

5.61000 = 561000 * 10^{-5}

1.1010 = 11010 * 10^{-4}

Then, Multiply both numbers

561000 * 10^{-5} * 11010 * 10^{-4}

First, rearrange

561000  * 11010 * 10^{-5} *  10^{-4}

6176610000 * 10^{-5} *  10^{-4}

From law of indices, 10^{-5} *  10^{-4} = 10^{-9};

So, we have

6176610000 * 10^{-9}

6.176610000

Because 5.61000 is given in 3 significant figures and 1.1010 is given in 4 significant figures, we approximate the result to 4 significant figures

This gives 6.177

Hence, 5.61000dg  * 1.1010dg = 6.177 dg^2

2.

12.0 m  ÷ 3.1415 m

Using proper notation

12.0 m ÷ 3.1415 m = \frac{12}{3.1415}

We start by representing 3.1415 using scientific notation

3.1415 = 31415 * 10^{-4}

By substitution;

\frac{12}{3.1415} = \frac{12}{31415 * 10^-4}

Take 10^{-4} to the numerator

\frac{12}{3.1415} = \frac{12 * 10^4}{31415}

From law of indices, 10^{4} = 10000

\frac{12}{3.1415} = \frac{12 * 10000}{31415}

Multiply

\frac{12}{3.1415} = \frac{120000}{31415}

Divide

\frac{12}{3.1415} = 3.81983129078

Because 12.0 is given in 2 significant figures and 3.1415 is given in 5 significant figures, we approximate the result to 5 significant figures

\frac{12}{3.1415} = 3.8198

anygoal [31]3 years ago
3 0

Answer:

1. 6.1767

2. 3.82

Step-by-step explanation:

Edge

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Answer:

2p + 2d = 39 ____________(1)

8p + 10d = 174.50 _________(2)

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Let the price of a bag of popcorn be p.

Let the price of a drink be d.

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2p + 2d = 39 ____________(1)

Ava spends a total of $174.50 on 8 bags of popcorn and 10 drinks. This implies that:

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