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alexira [117]
3 years ago
9

A patient receives 150 cc of medication in an IV drip over 4 hours. What is the rate per hour of this dose?

Mathematics
2 answers:
cupoosta [38]3 years ago
6 0
The average rate is 150 cc/4 hrs=37.5 cc per hour.
Andru [333]3 years ago
6 0
37.5 cc per hour you just have to divide 150cc/4 = 37.5 cc
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Step-by-step explanation:

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3 years ago
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Which equation is equivalent to log Subscript 3 Baseline (x + 5) = 2?
adell [148]

Answer:

The correct option is right-bracket squared 3 squared =x+5

Step-by-step explanation:

The equation is \log _{3}(x+5)=2

Option a: \log _{3}(x+5)=3^{2}

This is not possible because using logarithmic rule, if \log _{a} b=c then b=a^{c}

Hence, option a is not equivalent to \log _{3}(x+5)=2

Option b: \log _{3}(x+5)=2^{3}

This is not possible because using logarithmic rule, if \log _{a} b=c then b=a^{c}

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Option c: x+5=3^{2}

This is possible because using logarithmic rule, if \log _{a} b=c then b=a^{c}

Hence, option c is equivalent to \log _{3}(x+5)=2

Option b: x+5=2^{3}

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Hence, option b is not equivalent to \log _{3}(x+5)=2

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8 0
4 years ago
A necklace contains twenty-four spherical beads of silver, each having a radius of 0.5cm. The beads are to be coated, and the co
Inessa [10]

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15.3 x 10 = $153.

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4 years ago
To convert 4.1 kilometers to meters ,you would use the ratio 1,000meters/1 kilometer
boyakko [2]
4,100 meters. Hope it helped
7 0
4 years ago
Question regarding logarithms.
Eddi Din [679]

5^{x-2}-7^{x-3}=7^{x-5}+11\cdot5^{x-4}\\\\5^{x-2}-7^{x-2-1}=7^{x-2-3}+11\cdot5^{x-2-2}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\5^{x-2}-\dfrac{7^{x-2}}{7^1}=\dfrac{7^{x-2}}{7^3}+11\cdot\dfrac{5^{x-2}}{5^2}\\\\5^{x-2}-\dfrac{1}{7}\cdot7^{x-2}=\dfrac{1}{343}\cdot7^{x-2}+\dfrac{11}{25}\cdot5^{x-2}\\\\-\dfrac{1}{7}\cdot7^{x-2}-\dfrac{1}{343}\cdot7^{x-2}=\dfrac{11}{25}\cdot5^{x-2}-5^{x-2}\\\\\left(-\dfrac{1}{7}-\dfrac{1}{343}\right)\cdot7^{x-2}=\left(\dfrac{11}{25}-1\right)\cdot5^{x-2}

\left(-\dfrac{49}{343}-\dfrac{1}{343}\right)\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\\\\-\dfrac{50}{343}\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\qquad\text{multiply both sides by}\ \left(-\dfrac{25}{14}\right)\\\\\dfrac{50\cdot25}{343\cdot14}\cdot7^{x-2}=5^{x-2}\qquad\text{divide both sides by}\ 7^{x-2}\\\\\dfrac{25\cdot25}{343\cdot7}=\dfrac{5^{x-2}}{7^{x-2}}\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}

\dfrac{5^2\cdot5^2}{7^3\cdot7}=\left(\dfrac{5}{7}\right)^{x-2}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\\dfrac{5^4}{7^4}=\left(\dfrac{5}{7}\right)^{x-2}\\\\\left(\dfrac{5}{7}\right)^4=\left(\dfrac{5}{7}\right)^{x-2}\iff x-2=4\qquad\text{add 2 to both sides}\\\\\boxed{x=6}

6 0
3 years ago
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