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Solve the problems. Express your answers to the correct number of significant figures.

[tex]\[
\begin{array}{l}
\frac{2.31}{0.30}=\square \\
\left(2.08 \times 10^3\right) \times\left(3.11 \times 10^2\right)=\square \times 10^5
\end{array}
\][/tex]


Sagot :

Sure, let's solve the problems step by step.

1. First Calculation:

[tex]\[ \frac{2.31}{0.30} \][/tex]

To perform the division:

[tex]\[ \frac{2.31}{0.30} = 7.7 \][/tex]

The result is [tex]\( 7.7 \)[/tex] to the correct number of significant figures (two significant figures as in the divisor).

2. Second Calculation:

[tex]\[ \left(2.08 \times 10^3\right) \times \left(3.11 \times 10^2\right) \][/tex]

To perform the multiplication of numbers in scientific notation:

1. Multiply the coefficients (the base numbers):

[tex]\[ 2.08 \times 3.11 = 6.4488 \approx 6.45 \][/tex]

We round the result to three significant figures, because [tex]\(2.08\)[/tex] and [tex]\(3.11\)[/tex] both have three significant figures.

2. Add the exponents:

[tex]\[ 3 + 2 = 5 \][/tex]

Putting it all together:

[tex]\[ (2.08 \times 10^3) \times (3.11 \times 10^2) = 6.45 \times 10^5 \][/tex]

To summarize the solutions:

[tex]\[ \frac{2.31}{0.30} = 7.7 \][/tex]

[tex]\[ \left(2.08 \times 10^3\right) \times \left(3.11 \times 10^2\right) = 6.47 \times 10^5 \][/tex]

So the correct answers to fill in the boxes are:

[tex]\[ \boxed{7.7} \][/tex]

[tex]\[ \boxed{6.47} \times 10^5 \][/tex]