What is the solution? 1.00×10−153.9×10−7\frac{1.00 imes10^{-15}}{3.9 imes10^{-7}}

[Fill in the blank],
x 10[Fill in the blank],

Answer to y must have the correct number of significant figures.

Question

What is the solution? 1.00×10−153.9×10−7\frac{1.00	imes10^{-15}}{3.9	imes10^{-7}}

[Fill in the blank],
 x 10[Fill in the blank],

Answer to y must have the correct number of significant figures.

BlackTom AI · Accepted Answer

Here's how to approach this fill-in-the-blank problem, step by step, with attention to significant figures.
First, identify the given quantities: 1.00×10^−15 and 3.9×10^−7. The division you need is (1.00×10^−15) / (3.9×10^−7).
Next, divide the coefficients: 1.00 / 3.9 ≈ 0.256410... This coefficient will determine the leading digits of the result.
Now handle the powers of ten: 10^−15 divided by 10^−7 gives 10^(−15 + 7) = 10^(−8). So the raw result is 0.256410... × 10^(−8).
Convert to proper scientific notation: 0.256410... × 10^(−8) = 2.56410... × 10^(−9).
Consider significant figures: the number 1.00 has 3 significant figures, and 3.9 has 2 significant figures. When multiplying or dividing, the result should have as many significant figures as the factor with the fewest significant figures, which is 2 in this case.
Rounding 2.56410... × 10^(−9) to 2 significant figures gives 2.6 × 10^(−9). This corresponds to a coefficient of 2.6 and an exponent of −9 in scientific notation.
In the fill-in-the-blank format, you would place the two parts as: the first blank filled with 2.6, and the second blank filled with −9. The instruction about significant figures is consistent with keeping two significant digits in the coefficient while maintaining the proper scientific notation.
If you double-check, you can verify that 2.6 × 10^(−9) equals 2.6e−9, which matches the rounded result from 2.56410... × 10^(−9).

What is the solution? 1.00×10−153.9×10−7\frac{1.00\times10^{-15}}{3.9\times10^{-7}} [Fill in the blank], x 10[Fill in the blank], Answer to y must have the correct number of significant figures. Multiple fill-in-the-blank

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