Question

10 the body of a 154—pound person contains approximately $2 \times 10^{-1}$ milligrams of gold and $6 \times 10^1$ milligram this information, the number of milligrams of aluminum in the body is how many times the number of millig enter your answer in the box. 11 in scientific notation, $8,599,000,000 = 8.599 \times 10^x$. what is the value of $x$? a \\(\circ\\) 9 \\(\circ\\) 8 \\(\circ\\) 7 \\(\circ\\) 10 2 find the sum. write your answer in scientific notation. $(3.9 \times 10^4) + (2.5 \times 10^4)$ $6.4 \times 10^4$

Explanation:

Question 10 (assuming the missing part is about aluminum and gold, let's suppose we need to find how many times aluminum is of gold, with aluminum as \(6 \times 10^{1}\) and gold as \(2 \times 10^{-1}\))

Step1: Identify the values

Let \(A = 6 \times 10^{1}\) (aluminum) and \(G = 2 \times 10^{-1}\) (gold). We need to find \(\frac{A}{G}\).

Step2: Divide the coefficients and exponents

\(\frac{6 \times 10^{1}}{2 \times 10^{-1}}=\frac{6}{2}\times10^{1 - (-1)}\)

Step3: Calculate the result

\(\frac{6}{2}=3\) and \(1-(-1) = 2\), so \(3\times10^{2}=300\).

To write \(8,599,000,000\) in scientific notation \(a\times10^{x}\) where \(1\leq a<10\), we move the decimal point to get \(8.599\). We moved the decimal point 9 places (from the end of \(8,599,000,000\) to after 8). So \(x = 9\).

Step1: Factor out \(10^{4}\)

Since both terms have \(10^{4}\), we can factor it: \((3.9 + 2.5)\times10^{4}\)

Step2: Add the coefficients

\(3.9+2.5 = 6.4\)

Step3: Write in scientific notation

So the sum is \(6.4\times10^{4}\)

Answer:

\(300\)

Question 11