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adell [148]
2 years ago
5

How many mL of medication are needed to last 10 days if the dose of medication is 2.5 tsp TID (three times a day)?

Mathematics
1 answer:
rjkz [21]2 years ago
7 0

Answer:

369.7 mL of medication

Step-by-step explanation:

How many mL of medication are needed to last 10 days if the dose of medication is 2.5 tsp TID (three times a day)?

From the above question,

The dosage of the medication =

2.5 tsp 3 times a day

= 2.5 × 3 = 7.5 tsp per day.

Since

1 day = 7.5 tsp

10 days = x tsp

Cross Multiply

x = 10 × 7.5 tsp

x = 75 tsp of medication for 10 days.

Step 2

It is important to note that:

1 tsp = 4.929 mL

75 tsp = x mL

Cross Multiply

x = 75 × 4.929 mL

x = 369.669 mL of medication

Approximately = 369.7 mL of medication

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Step-by-step explanation:

I think thats the answer,

4 0
2 years ago
754,863 rounded to the nearest ten thousands place
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754,863 rounded to the nearest ten thousands is 750,000
7 0
3 years ago
The parallelogram has a base of 8 cm and corresponding height of 4 cm. The parallelogram is changed proportionally and is dilate
sukhopar [10]

Answer:

The area of the new reduced parallelogram after dilation is 8 cm^2

Step-by-step explanation:

Mathematically, the area of a parallelogram = b * h

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After dilation by a factor of 1/2, the base of the parallelogram becomes 1/2 * 8 = 4cm while the height becomes 1/2 * 4 = 2cm

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8 0
3 years ago
The equation x+53=15 can be solved by performing two operations on both sides.
Wewaii [24]
Hey!


You're right! This equation can definitely be solved by performing two operations on both sides. Let me show you how it's done.

First, let's write out our equation.

<span><em>Original Equation :</em>
</span>x + 53 = 15

Now that that's done, we'll be performing two operations on both sides. The operation we'll want to do is subtracting both sides of the equation by 53. We do this to get x on its own.

<em>Original Equation :</em>
x + 53 = 15

<em>New Equation {Added Subtract 53 to Both Sides} :</em>
x + 53 - 53 = 15 - 53

Now we have to solve the equation. Let's do the left side first.

<em>Left Side of the Equation :</em>
x + 53 - 53 

<em>Left Side of the Equation {Solved} :</em>
x

Now, we'll solve the right side of the equation.

<em>Right Side of the Equation :</em>
15 - 53

<em>Right Side of the Equation {Solved} :</em>
-38

Now we can put both solutions to both sides of the equation together.

<em>New Equation :</em>
x = -38

Since this cannot be simplified any farther, this is our final answer. And that's it!

<em>So, now we know that in the equation x + 53 = 15,</em>  x = -38

Hope this helps!


- Lindsey Frazier ♥
3 0
3 years ago
How do you solve this? My algebra 2 teacher taught it to us in class but it still confuses me! Thank you!
nataly862011 [7]

\bf \cfrac{\frac{1}{8}-\frac{1}{2x}}{\frac{1}{12x}-\frac{1}{3x^2}} \begin{array}{llll} \leftarrow \textit{our LCD here will just be }8x\\[0.9em] \leftarrow \textit{our LCD down here will be }12x^2 \end{array}

\bf \cfrac{~~\frac{(x)1-(4)1}{8x}~~}{\frac{(x)1-(4)1}{12x^2}}\implies \cfrac{~~\frac{x-4}{8x}~~}{\frac{x-4}{12x^2}}\implies \cfrac{~~\begin{matrix} x-4 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{\underset{2}{~~\begin{matrix} 8x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}\cdot \cfrac{\stackrel{3x}{~~\begin{matrix} 12x^2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}{~~\begin{matrix} x-4 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }\implies \cfrac{3x}{2}

8 0
3 years ago
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