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Helga [31]
1 year ago
7

In spinal motion restriction of the adult, which action demonstrates correct stabilization technique

SAT
1 answer:
nirvana33 [79]1 year ago
7 0

In spinal motion restriction of the adult, to maintain the client's nose and umbilicus aligned at all times demonstrates the correct stabilization technique.

<h3>What is spinal motion restriction?</h3>

Spinal motion restriction is a strategy used to maintain spine alignment during spinal injury.

This technique is fundamental to avoiding severe spinal injury after accidents or traumatic situations.

In conclusion, in spinal motion restriction of the adult, to maintain the client's nose and umbilicus aligned at all times demonstrates the correct stabilization technique.

Learn more about spinal motion here:

brainly.com/question/20215668

#SPJ1

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A grocery store has two recycling machines
Thepotemich [5.8K]

Answer is 112%

kinda skeptical about my answer but i am going to give it my best shot, sorry if i get it wrong.

In percent-type questions I always go by this formula:

not a %                            %

original number       :     100%

new number            :      new %

original number × new %  =  new number × 100%

here, the original numbers are 240 and 180 and the new % is +30% and -10% now let plastic be P and metal be M and let's elaborate this a bit further

plastic:

the first recycling machine took in 240  plastic bottles while the second took in 30% <u>more</u> plastic bottles, what we have to do here is to find out how much bottles the second machine took in as a form of integer, which is the new number; let that new number be X, let's plug this in the formula

240:100+30\\X:130 \\

notice: 130 is the new %

240*130 = 100X\\31200 = 100X\\312=X

the second machine took in 312 plastic bottles, let's save this answer for later

metal:

the first recycling machine took in 180 metal cans while the second took in 10% <u>less</u> plastic bottles, we do the same thing and find the new number, this time let it be Y:

180:100-10\\Y=90

90 is the new % here; also, notice how sometimes you either add or subtract percents, depending on what the problem demands.

180*90=100Y\\16200=100Y\\162=Y

Now we can conclude that the first recycling machine took in 240  plastic bottles and 180 metal cans and the second took in 312 plastic bottles and 162 metal cans.

here comes the hardest bit to apprehend and the final part of this problem, "the second machine recycled what percent (%) more items than the first?"

in other words, how much percent did the second machine recycle more.

to find that percent, first we need to add the plastic and metal "items" of both machines individually and turn them into percents: let the first machine be A and second machine be B:

A: 240+180=420

B: 312+162=474

again, using the same formula to find the new percent, let that percent be Z; plugging both A and B respectively:

420:100\\474:Z

420Z=474*100\\420Z=47400\\Z= 112.857≈ 112% more items than the first

now if you check the rounded up 112 and plug it back in to make sure of your answer: 420Z=47400 you end up getting 47040, since it's rounded up, but when you plug the full 112.857 you get 47399.94, closest to 47400. Tip: when you plug in any value for check up, don't use the rounded up one.

At last, those steps are actually fast, especially with that formula it's so important. the explanation seems intimidating so read it slowly and take your time. if you have any other questions, feel free to ask. also, let me know if you know a faster method to solve this question. <3

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