The quotient of x and 2.

Mathematics

1 answer:

jarptica [38.1K]4 years ago

6 0

whattttt???? This is too hardd

You might be interested in

For three consecutive years, Sam invested some money at the start of the year. The first year, he invested x dollars. The second

faust18 [17]

$\bf \stackrel{\textit{Sam invested the first year}}{x} \qquad \qquad \stackrel{\textit{2nd year is \$2,000 less than \underline{x}}}{x-2000} \\\\\\ \stackrel{\textit{3rd year is \$1,000 more than \underline{x}}}{x+1000}\qquad \qquad \boxed{\stackrel{\textit{Sam total investment}}{(x)+(x-2000)+(x+1000)}}\\\\ -------------------------------\\\\ \stackrel{\textit{Sally invested the first year \$1000 less than \underline{x}}}{x-1000}$

$\bf \stackrel{\textit{2nd year is \$1500 less than \underline{2x}}}{2x-15000}\qquad \qquad \stackrel{\textit{3rd year is \$1400 more than \underline{x}}}{x+1400} \\\\\\ \boxed{\stackrel{\textit{Sally total investment}}{(x-1000)+(2x-1500)+(x+1400)}}$

$\bf \stackrel{\textit{Sam total investment}}{(x)+(x-2000)+(x+1000)}=\stackrel{\textit{Sally total investment}}{(x-1000)+(2x-1500)+(x+1400)} \\\\\\ 3x-1000=4x-1100\implies 100=x\impliedby \textit{Sam invested the first year} \\\\\\ \stackrel{\textit{Sally invested last year}}{x+1400}\implies 10+1400\implies 1500$

4 0

4 years ago

Which comparison is correct?

r-ruslan [8.4K]

Option B: $4.5 \times 10^{-6}>3.9 \times 10^{-9}$ is the correct answer.

Explanation:

Option A: $7.1 \times 10^{8}>1.2 \times 10^{9}$

Simplifying, we get,

$710000000>1200000000$

Since, the value of $1.2 \times 10^{9}$ is greater than $7.1 \times 10^{8}$ , the comparison $7.1 \times 10^{8}>1.2 \times 10^{9}$ is false.

Hence, Option A is not the correct answer.

Option B: $4.5 \times 10^{-6}>3.9 \times 10^{-9}$

Simplifying, we get,

$0.0000045>0.0000000039$

Since, the value of $4.5 \times 10^{-6}$ is greater than $3.9 \times 10^{-9}$ , the comparison $4.5 \times 10^{-6}>3.9 \times 10^{-9}$ is true.