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4 years ago

5

Describe a relationship in which one quantity increases as another quantity decreases

1 answer:

Morgarella [4.7K]4 years ago

4 0

Inversely proportional

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A 3mL unit dose of levalbuterol HCl contains 0.63mg of active ingredient. If 2 unit doses are used, how many mg of active ingred

Sergio039 [100]

Answer: 0.42 mg

Step-by-step explanation:

Given : A 3mL unit dose of levalbuterol HCl contains 0.63mg of active ingredient.

Then , the amount of active ingredient in 1 ml dose of levalbuterol HCl = $\dfrac{0.63}{3}$

$=\dfrac{63}{300}=\dfrac{21}{100}=0.21\ mg$

Hence, the amount of active ingredient in 2 ml dose of levalbuterol HCl will be :-

$2\times0.21=0.42\ mg$

Hence, the amount of active ingredient in 2 ml dose of levalbuterol HCl =0.42 mg

6 0

4 years ago

PLEASE HELP ASAP<br> What is the value of x?

dedylja [7]

We know that a triangle is 180 degrees.

$102+39=141$

$180-141=39$

x=39

4 0

3 years ago

Read 2 more answers

Which ordered pair is a solution to the system of linear equations 1/2x-3/4y=11/60 and 2/5x+1/6y=3/10

natka813 [3]

ANSWER

$( \frac{2}{3} , \frac{1}{5} )$

EXPLANATION

The first equation is

$\frac{1}{2} x - \frac{3}{4} y = \frac{11}{60} ...(1)$

The second equation is

$\frac{2}{5} x + \frac{1}{6} y = \frac{3}{10} ...(2)$

We want to eliminate y, so we multiply the first equation by

$\frac{4}{5}$

$\frac{4}{5} \times \frac{1}{2} x - \frac{4}{5} \times \frac{3}{4} y = \frac{11}{60} \times \frac{4}{5}$

$\frac{2}{5} x - \frac{3}{5} y = \frac{11}{75} ...(3)$

We now subtract equation (3) from (2)

$(\frac{2}{3} x - \frac{2}{3} x )+ ( \frac{1}{6} y - - \frac{3}{5}y ) =( \frac{3}{10} - \frac{11}{75} )$

$\frac{1}{6} y + \frac{3}{5}y =\frac{3}{10} - \frac{11}{75}$

$\frac{23}{30}y = \frac{23}{150}$

Multiply both sides by

$\frac{30}{23}$

$\implies \: \frac{30}{23} \times \frac{23}{30}y= \frac{23}{150} \times \frac{30}{23}$

$\implies \: y = \frac{1}{5}$

Substitute into the first equation to solve for x .

$\frac{1}{2} x - \frac{3}{4} \times \frac{1}{5} = \frac{11}{60}$

Multiply to obtain

$\frac{1}{2} x - \frac{3}{20} = \frac{11}{60}$

$\frac{1}{2} x = \frac{11}{60} + \frac{3}{20}$

$\frac{1}{2} x = \frac{1}{3}$

Multiply both sides by 2.

$2 \times \frac{1}{2} x =2 \times \frac{1}{3}$

$x = \frac{2}{3}$

The solution is

$( \frac{2}{3} , \frac{1}{5} )$

5 0

3 years ago

Read 2 more answers

Write a system of linear equations for the graph below

Goshia [24]

Answer:

$\left \{{{y=3-\frac{x}{3} }\atop { y=-2x-2}} \right.$

Step-by-step explanation:

$(-3,4),(0,3)\\y=3-\frac{x}{3} \\(-3,4),(0,-2)\\\\y=-2x-2$

7 0

3 years ago

Find the greatest common factor of the terms of the polynomial.<br> ♡<br> 8k^4 + 8k^2– 4k

9966 [12]

Answer:

4k

Step-by-step explanation:

6 0

3 years ago