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Mathematics

2 answers:

dangina [55]2 years ago

6 0

Answer:

$\frac{0.3x - 2.1}{3} = 0.4 \\ 0.3x - 2.1 = 0.4 \times 3 \\ 0.3x - 2.1 = 0.12 \\ 0.3 x = 0.12 + 2.1 \\ 0.3x = 2.22 \\ x = \frac{2.22}{0.3} \\ x = 7.4$

ollegr [7]2 years ago

5 0

Answer:

$\frac{0.3x - 2.1}{3} = 0.4 \\ 0.3x - 2.1 = 0.4 \times 3 \\ 0.3x - 2.1 = 1.2 \\ 0.3x = 1.2 + 2.1 \\ 0.3x = 3.3 \\ x = \frac{3.3}{0.3} \\ \boxed{x = 11}$

<u>x = 11</u> is the right answer.

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Reggie was looking at a map of the school the scale shows that 1 inch equals 150 ft if the distance between his classroom in the

topjm [15]

Answer:

the actual distance is 1800 ft.

Step-by-step explanation:

Proportion states that the two fractions or ratios are equal.

Let x be the actual distance.

As per the statement:

Reggie was looking at a map of the school the scale shows that 1 inch equals 150 ft.

⇒1 inch = 150 ft.

if the distance between his classroom in the cafeteria measures 12 in on the map.

by definition of proportions we have;

$\frac{1}{150} = \frac{12}{x}$

By cross multiply we have;

$x = 12 \cdot 150 = 1,800$ ft

Therefore, the actual distance is 1800 ft.

4 0

3 years ago

Read 2 more answers

How many milliequivalents of sodium chloride are contained with 3 L of normal saline?

AVprozaik [17]

Answer:

466mEq

Step-by-step explanation:

First, we need to know the concentration of NaCl in a normal saline solution, this is by definition 0.9%, meaning we have 0.9g of NaCl per 100ml of solution, we want to know how much NaCl we have in 3L (3000ml):

$3000ml*\frac{0.9g}{100ml}=27g=27000mg$

So, we have 27000mg in 3L of normal saline solution.

Now, acording to our milliequivalent (mEq) equation ( $mEq=\frac{mg}{pE}$ ) where pE is de molecular mass of NaCl divided by their charges, in this case 1: