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2 years ago

14

A patient has a twenty-centimeter incision along the abdomen that must have a topical antibiotic ointment applied for every dres

sing change. How should the amount of medication to apply at home in inches be written?

1 answer:

rodikova [14]2 years ago

4 0

Answer:

The amount of medication to be applied is approximately 7.87 inches

Step-by-step explanation:

The given parameters are;

The length of the incision the patient has = 20 cm

The length of the application of the antibiotic = The length of the incision the patient has

∴ The length of the application of the antibiotic = 20 cm

The amount of medication to be applied = 20 cm = 7.874016 inches ≈ 7.87 inches

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Evaluate composite functions

Vlada [557]

Answer:

2

Step-by-step explanation:

Step 1: Write out equation g(f(a))

1/√a

Step 2: Plug in 0.25 for <em>a</em>

1/√0.25

You answer should be 2

5 0

2 years ago

Last week, it rained g inches. This week, the amount of rain decreased by 5%. Which expressions represent the amount of rain tha

dybincka [34]

Answer:

The correct answers are:

0.95g

g - 0.05g

(1 - 0.05)g

Step-by-step explanation:

Amount of last week's rain = g inches

this week = 5% decrease on last week's amount

5% of last week's amount = 5/100 × g = 0.05g

∴ This week's amount of rain = 0.05g decrease

∴ Amount of rain this week = (Amount of rain last week) - 0.05g

∴ Amount of rain this week = g - 0.05g

This can be simplified by factorizing the common term "g"

g - 0.05g = g(1 - 0.05) = 0.95g

4 0

2 years ago

I’m working on 3 and 4 only maybe if I can have some explanation so I can understand and do more of these next week.

guapka [62]

4.b.

Answer: See below.

Step-by-step explanation:

<h2><u>For the equation f(x) = 2x</u></h2>

3.a. f(6) means use x = 6 in the equation f(x) = 2x

so f(6) would be f(6)= 2(6)

<u>f(6) = 12</u>

3.b. f(-11) = 2(-11)

<u>f(-11) = -22</u>

3.c. f(2.75) = 2(2.75)

<u>f(2.75) = 5.5</u>

3.d. This is turned around. We are told f(x)=20, so what would x need to be for f(x) to be 20? Since f(x) = 2x, we can say 20 = 2x. Therefore x = 10

f(10) = 20

<u>The rest of (3) are solved in the same fasion.h</u>

<u></u>

<h2><u>For the equation f(x)= 5x+50</u></h2>

4.a. f(7) = 5(7)+50

<u>f(7) = 85</u>

4.b. f(-12)

f(-12) = 5*(-12)+50

<u>f(-12) = -60</u>

<u></u>

Continue in the same fashion for these types of problems.

7 0

2 years ago

algol [13]

Answer:

1:350 i think

Step-by-step explanation:

i could be wrong

7 0

2 years ago

Read 2 more answers

Please solve this<br> (Rational numbers 8th grade)

pochemuha

Question # 1

Answer:

$\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}=-\frac{3}{20}$

Step-by-step explanation:

Given the expression

$\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}$

$=-\frac{2}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}$ ∵ $\frac{-2}{3}\times \frac{3}{5}=-\frac{2}{5}$

$=-\frac{2}{5}+\frac{1}{4}$ ∵ $\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}=\frac{1}{4}$

$\mathrm{Least\:Common\:Multiplier\:of\:}5,\:4:\quad 20$

$=-\frac{8}{20}+\frac{5}{20}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{-8+5}{20}$

$\mathrm{Add/Subtract\:the\:numbers:}\:-8+5=-3$

$=\frac{-3}{20}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}$

$=-\frac{3}{20}$

Therefore,

$\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}=-\frac{3}{20}$

Question # 2

Answer:

$\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}=-\frac{5}{28}$

Step-by-step explanation:

Given

$\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}$

$=-\frac{6}{35}-\frac{1}{6}\times \frac{3}{2}\times \frac{2}{5}\times \frac{1}{14}$ ∵ $\frac{2}{5}\times \frac{-3}{7}=-\frac{6}{35}$

$=-\frac{6}{35}-\frac{1}{140}$ ∵ $\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}=\frac{1}{140}$

$\mathrm{Least\:Common\:Multiplier\:of\:}35,\:140:\quad 140$

$=-\frac{24}{140}-\frac{1}{140}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{-24-1}{140}$

$\mathrm{Subtract\:the\:numbers:}\:-24-1=-25$

$=\frac{-25}{140}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}$

$=-\frac{25}{140}$

$\mathrm{Cancel\:the\:common\:factor:}\:5$

$=-\frac{5}{28}$

Therefore,

$\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}=-\frac{5}{28}$

4 0

3 years ago