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2 years ago

13

a newborn is treated for pulmonary valve stenosis; stretching of the valve opening is accomplished via a percutaneous balloon pu

lmonary valvuloplasty. what is the root operation?

1 answer:

r-ruslan [8.4K]2 years ago

3 0

Answer:

dilation

Step-by-step explanation:

When the pulmonary valve does not work properly, it can interfere with blood flow from the heart to the lungs, as well as force the heart to work harder to carry the blood that is needed to the rest of the body. Some children have heart conditions present at the time of birth and may require repair or replacement of the pulmonary valve, this option has a lower risk of infection, preserves the strength and functioning of the valve, and eliminates the need to take medication.

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PLEASE HELP!!!!!

Zielflug [23.3K]

Question #1

Part A:
The y-intercept can be found when x = 0. If you look at your table, when x = 0, y = 5. So the y-intercept is 5.

Part B:
$\sf Slope = \frac{27-5}{1-0} = \frac{22}{1} = \boxed{22}$
The slope is 22.

Part C:
y = mx + b
y = 22x + 5

We are given 225 as the range, or in place of y.
225 = 22x + 5
220 = 22x
x = 10

The domain is 10.

Question #2

Part A:

(2,255)
(5,480)

Standard form is Ax + By = C

$\sf Slope = \frac{y_2-y_1}{x_2-x_1} = \frac{480-225}{5-2} = \frac{255}{3} = \boxed{85}$

Let's plug this into this form first:
$y - y_1 = m(x-x_1)\\\\y -225 = 85(x-2)\\\\y - 225 = 85x - 170\\\\y = 85x + 55$

Now, let's make it into Standard Form.
$y = 85x + 55\\\\y - 85x + 55\\\\ -85x + y = 55\\\\ -1(-85x +y) = -1(55)\\\\\boxed{85x - y = -55}$
What, which is in the box, is your final answer. :)

Part B:
Function notation simply means replacing y with f(x).
We had y = 85x + 55
So your answer is:
$\boxed{f(x) = 85x + 55}$

Part C:
Using the final answer which we got in Part A, we would know that the y-intercept is (0,55) and the x-intercept is (-55/85, 0). We would plot these 2 points, and then draw a line between them. :)

5 0

3 years ago

5.) Find the area of the triangle in bold.

trapecia [35]

Answer:

the answer is 0.98

Step-by-step explanation:

if it not correct plz correct me

4 0

2 years ago

Read 2 more answers

A solution initially contains 200 bacteria. 1. Assuming the number y increases at a rate proportional to the number present, wri

GuDViN [60]

Answer:

1. $\frac{dy}{dt}=ky$

2.543.6

Step-by-step explanation:

We are given that

y(0)=200

Let y be the number of bacteria at any time

$\frac{dy}{dt}$ =Number of bacteria per unit time

$\frac{dy}{dt}\proportional y$

$\frac{dy}{dt}=ky$

Where k=Proportionality constant

2. $\frac{dy}{y}=kdt$ ,y'(0)=100

Integrating on both sides then, we get

$lny=kt+C$

We have y(0)=200

Substitute the values then , we get

$ln 200=k(0)+C$

$C=ln 200$

Substitute the value of C then we get

$ln y=kt+ln 200$

$ln y-ln200=kt$

$ln\frac{y}{200}=kt$

$\frac{y}{200}=e^{kt}$

$y=200e^{kt}$

Differentiate w.r.t

$y'=200ke^{kt}$

Substitute the given condition then, we get

$100=200ke^{0}=200 \;because \;e^0=1$

$k=\frac{100}{200}=\frac{1}{2}$

$y=200e^{\frac{t}{2}}$

Substitute t=2

Then, we get $y=200e^{\frac{2}{2}}=200e$

$y=200(2.718)=543.6=543.6$

e=2.718

Hence, the number of bacteria after 2 hours=543.6

4 0

3 years ago

Need help on #5, please help :)

Gre4nikov [31]

Okay i answered by comment but now i can make an official answer here:

a) i got 5,828.6 grams / about 13lb

b) i got 273,391.48$

hope this helps!

6 0

3 years ago

Jarrett's puppy weighed 3 3/4 ounces at Birth. At 1 week old, the puppy weighed 5 1/8 ounces. At 2 weeks old, the puppy weighed

Ad libitum [116K]

Answer: The answer is $7\dfrac{7}{8}$ ounces.

Step-by-step explanation: Given in the question the weight of Jarrett's puppy at birth, one week old and 2 weeks old are as follows:

$b_0=3\dfrac{3}{4}=\dfrac{15}{4}~\textup{ounces},\\\\\\b_1=5\dfrac{1}{8}=\dfrac{41}{8}~\textup{ounces},\\\\\\b_3=6\dfrac{1}{2}=\dfrac{13}{2}~\textup{ounces}.$

We can see that

$b_1-b_0=b_2-b_1=\dfrac{11}{8}.$

So, the weight of the baby will make an arithmetic progression with first term 'a' and common difference 'd' as follows:

$b_0=\dfrac{15}{4},~~d=\dfrac{11}{8}.$

Thus, the weight of the puupy after 3 weeks will be

$b_3=b_0+3\times d=\dfrac{15}{4}+3\times\dfrac{11}{8}=\dfrac{30+33}{8}=\dfrac{63}{8}=7\dfrac{7}{8}~\textup{ounces}.$

5 0

3 years ago

Read 2 more answers