If you always follow the same five steps to get ready for school, then you are following an algorithm.

Engineering

2 answers:

raketka [301]2 years ago

4 0

Answer:

hmm true

Explanation:

becuase i think alogorthm is wht u do regurkry if i helped mark me brailist

DochEvi [55]2 years ago

3 0

Answer:

true

Explanation:

an algorithm is like your basic way of doing something always in the same way.

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Joseph wants to practice architecture. Which compulsory assessment administered by NCARB does he need to complete?

Hunter-Best [27]

Answer:

Architect Registration Examination (ARE)

Explanation:

In this scenario, Joseph wants to practice architecture. A compulsory assessment administered by National Council of Architectural Registration Boards (NCARB) which he has to complete is an Architect Registration Examination.

An Architect Registration Examination (ARE) refers to the professional licensure examination to be taken by anyone who intends to practice architecture in the United States of America, Puerto Rico, Guam, Canada, US Virgin Islands.

The main purpose of the Architect Registration Examination is to assess an architect's knowledge, skills, and abilities on architectural best practices, procedures and services. Also, it focuses on areas relating to a building's safety, soundness and health impact on the habitants. Therefore, ARE is a prerequisite for practicing architecture across the United States of America jurisdiction.

Once an architect passes the examination, he or she would be given a license to practice architecture.

<em>Hence, for Joseph to practice architecture he must take the Architect Registration Examination developed and administered by National Council of Architectural Registration Boards (NCARB). </em>

8 0

3 years ago

A commuter train traveling at 50 mi/h is 3 mi from a station. The train then decelerates so that its speed is 15 mi/h when it is

jonny [76]

Answer:

a) $t = 277.477\,s\,(4.625\min)$ , b) $v_{f} = 0\,\frac{mi}{h}$ , c) $a = -0.128\,\frac{ft}{s^{2}}$

Explanation:

a) The deceleration experimented by the commuter train in the first 2.5 miles is:

$a=\frac{[(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}-[(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (2.5\,mi)\cdot (\frac{5280\,ft}{1\,mi} )}$

$a = -0.185\,\frac{ft}{s^{2}}$

The time required to travel is:

$t = \frac{(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )-(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )}{-0.185\,\frac{ft}{s^{2}} }$

$t = 277.477\,s\,(4.625\min)$

b) The commuter train must stop when it reaches the station to receive passengers. Hence, speed of train must be $v_{f} = 0\,\frac{mi}{h}$ .

c) The final constant deceleration is:

$a = \frac{(0\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )-(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )}{(2.875\,min)\cdot (\frac{60\,s}{1\,min} )}$