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melomori [17]
2 years ago
6

A 34-pound child with epilepsy is prescribed Tegretol (carbamazepine) 15 mg/kg/day, taken four doses. Tegretol suspension is 100

mg/5 mL. What’s the volume of one dose of medication?
Mathematics
1 answer:
gayaneshka [121]2 years ago
8 0

Answer:

ummmm QwQ this is to hard T~T

Step-by-step explanation:

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Enter the sum of the numbers as the product of their GCF and another sum. 42 + 30
sesenic [268]

Answer:

the answer would be 72

Step-by-step explanation:

7 0
2 years ago
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Reposting this because no body gave me the equation. An equation would be appreciated :)
o-na [289]

bring the like terms together,

\frac{6q}{5}  +  \frac{4q}{5}  \leqslant 28 + 3 + 3 - 3

\frac{10q}{5}  \leqslant 31

2q \leqslant 31

q \leqslant  \frac{31}{2}

3 0
3 years ago
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Jackson bought 555 ounces of raisins for \$4$4dollar sign, 4. How much do raisins cost per ounce?
stepan [7]

Answer:

$0.80/oz

Step-by-step explanation:

$4 for 5 ounces

unit cost = price/weight = $4/(5 oz.) = $0.80/oz

8 0
3 years ago
The following table shows a proportional relationship between m and n
slavikrds [6]

Answer:

m = 7 n ( the '=' is meant to be the proportional sign)

Step-by-step explanation:

21 ÷ 3 = 7

35 ÷ 5 = 7

56 ÷ 8 = 7

7 0
3 years ago
Read 2 more answers
HELP PLEASE!! I DONT UNDERSTAND!!!!!!!!!! THANKS SO MUCH
8090 [49]

Hello, please consider the following.

We will multiply the numerator and denominator by

4+\sqrt{6x}

to get rid of the root in the denominator.

First of all, we cannot divide by 0, right? So, we need to make sure that the denominator is different from 0.

4-\sqrt{6x} =0\sqrt{6x}=4\\\\\text{Take the square}\\\\6x=4^2=16\\\\x=\dfrac{16}{6}=\dfrac{8}{3}

We need to take any x real number different from 8/3 then and simplify the expression.

Let's do it!

\begin{aligned}\dfrac{4}{4-\sqrt{6x}}&=\dfrac{4(4+\sqrt{6x})}{(4+\sqrt{6x})(4-\sqrt{6x})}\\\\&=\dfrac{4(4+\sqrt{6x})}{(4^2-\sqrt{6x}^2)}\\\\&=\dfrac{4(4+\sqrt{6x})}{(16-6x)}\\\\&=\dfrac{2(4+\sqrt{6x})}{(8-3x)}\\\\&\large \boxed{=\dfrac{8+2\sqrt{6x}}{8-3x}}\end{aligned}

Thank you

8 0
3 years ago
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