Given: 3x < -6.

1 answer:

7 0

$3x\ \textless \ -6 :$

Divide both sides by 3 :

$\frac{3x}{3} = \frac{-6}{3}$

$x\ \textless \ -2$

Answer A

<span>{x | x < -2}
</span>
hope this helps!

You might be interested in

A chemist is using 362 milliliters of a solution of acid and water. If 12.1% of the solution is acid, how many milliliters of ac

PSYCHO15rus [73]

Answer:

The answer to your question is 43.8 mL of acid.

Step-by-step explanation:

Data

Total volume = 362 ml

12.1 % is acid

The Volume of acid = ?

Process

1.- Use the rule of three and cross multiplication to solve this problem.

362 ml ------------------- 100%

x ------------------- 12.1 %

x = (12.1 x 362) / 100

x = 4380.2 / 100

x = 43.802 ml

2.- Round your answer to the nearest tenth.

43.802 only consider the first decimal

43.8 mL

3.- Conclusion

A 12.1% solution of acid has 43.8 mL of acid.

6 0

3 years ago

Read 2 more answers

A gardener works for 13 hours during a week and charges $169 how much does the gardener charge for each hour?

bonufazy [111]

Answer:

You have to divide to unit price, 169/13 is equal to 13!

Step-by-step explanation:

7 0

2 years ago

Write an expression for the 12th partial sum of the series 3/2+7/3+19/6+... using summation notation

lapo4ka [179]

Answer:

$S_{12}=\sum_{i=1}^{12} [\frac{3}{2}+(i-1)\times \frac{5}{6}]$

$S_{12}=73$

Step-by-step explanation:

$First\ term\ of\ the\ series(a_1)=\frac{3}{2}\\\\Second\ term\ of\ the\ series(a_2)=\frac{7}{3}\\\\Third\ term\ of\ the\ series(a_3)=\frac{19}{6}\\\\a_2-a_1=\frac{7}{3}-\frac{3}{2}=\frac{5}{6}\\\\a_3-a_2=\frac{19}{6}-\frac{7}{3}=\frac{5}{6}\\\\Hence\ it\ is\ an\ Arithmetic\ Series\ with\ first\ term=\frac{3}{2}\ and\ constant\ difference=\frac{5}{6}$

$a_1=\frac{3}{2}+0\times \frac{5}{6}\\\\a_2=\frac{3}{2}+1\times \frac{5}{6}\\\\a_3=\frac{3}{2}+2\times \frac{5}{6}\\\\.\\.\\.\\a_n=\frac{3}{2}+(n-1)\times \frac{5}{6}\\\\S_n=a_1+a_2+a_3+......+a_n\\\\S_n=(\frac{3}{2}+0\times \frac{5}{6})+(\frac{3}{2}+1\times \frac{5}{6})+(\frac{3}{2}+2\times \frac{5}{6})+....+(\frac{3}{2}+[n-1]\times \frac{5}{6})\\\\S_n=\sum_{i=1}^n [\frac{3}{2}+(i-1)\times \frac{5}{6}]\\\\S_n=(\frac{3}{2}+\frac{3}{2}+\frac{3}{2}+...n\ times)+\frac{5}{6}(1+2+3+4+...+(n-1))\\\\$