Answer:
The most likely galaxy type to be identifiable regardless of orientation is: Irr
Explanation:
The Irr galaxies don't have a discernable or usual shape; that is why it is relatively easy to identify.
When we talk about E type galaxies, this statement proves itself by the way the cumulus of stars compounds the galaxy. The elliptical galaxies have the form of ellipses, with a reasonable distribution of stars. The degree of eccentricity is the number that complements the E letter; that's why E0 galaxies are almost spherical, while E7 is considerably elongated.
SBc, SBa galaxies are spiral; this means it can be flat in some angles difficulting their identification process; in this case, the last letter means the way the arms display their form, with "c" having a vague form and "a" well-defined arms. That's why in some angles can be mistreated as another type of galaxy.
I guess the correct answer is plant system pathway
Thе plant systеms pathway includеs οccupatiοns rеlatеd tο grοwing fοοd, fееd, and fibеr crοps, and thе study οf plantsand thеir grοwth tο hеlp prοducеrs mееt cοnsumеr dеmand whilе cοnsеrving natural rеsοurcеs and maintaining thе еnvirοnmеnt.
In his work for a new company, Byron found a flower material that he could use to manufacture dresses. In his career, Byron is most likely focused on the plant system pathway.
Answer:
optical sensors
Explanation:
optical sensors measure the amount of gases like carbon dioxide and carbon monoxide in the air
Answer: civil engineer
Explanation:
Based on the information given, the type of engineer that would be identified as essential to the success of the project would be the civil engineer.
Civil engineers are the engineers that are in charge of the planning and overseeing building and infrastructure construction. They plan and monitor constructions involving bridges, road, houses, power plants etc.
Answer:
please find the attachment of the flowchart.
Explanation:
In this question, a start block is used to start the program, in the parallelogram box we input the value from the user end and in the diamond box, we check the input is not equal to 0. In the next diamond box, it checks the given input value and print value, and at the last, we stop the code.
Please find the program and its output in the attached file.
