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Nitella [24]
3 years ago
12

How does the start function make your programs easier to understand?

Engineering
1 answer:
ad-work [718]3 years ago
7 0

Explanation:

All of the commands you wish to execute must go inside of the start function, between the curly braces.

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According to the eNotes, a program that eliminates sales and promotions in an effort to minimize the bullwhip effect would be ca
Ilia_Sergeevich [38]

Options: True or false

Answer:True, it will be called CRMWRONGEDLP.

Explanation:eNotes was founded in 1998 by Brad Satoris and Alexander Bloomingdale, to enable and enhance the ability of students to solve assignments as given and other home works,eNote is also used by students to prepare effectively and efficiently for examinations.

Since its launch eNote has been a great source of educational materials for students who have gained a lot in the use of its materials.

bullwhip effect is a concept in supply chain, where forcasts causes the supply chain to be inefficient, it usually starts from retailers who raise concern of high demands.

CRMWRONGEDLP, is the program that minimizes the bullwhip effects according to eNote.

7 0
3 years ago
Which one is dependent variable?
GREYUIT [131]

Answer:

The dependent variable is MEDV - Median value of owner-occupied homes in $1000's

Explanation:

The median value of the house has to be predicted, based on its properties and neighborhood properties, this can be done by using a linear regression model.

The dependent variable in Machine Learning is the output variable that we want to predict.

Therefore, according to the question given "MEDV" is the dependent variable.

8 0
3 years ago
A cantilever beam AB of length L has fixed support at A and spring support at B.
atroni [7]

The force in the spring will be F =\dfrac{KPl^3}{3EI}.

The deflection of the beam will be \rho = 0.15(\dfrac{KPL^3}{3EI})

<h3>What is a cantilever beam?</h3>

A rigid, horizontally extending structural member known as a cantilever is supported at only one end. Typically, it extends from a solidly affixed flat vertical surface, such as a wall.

Given that:-

  • A cantilever beam AB of length L has fixed support at A and spring support at B.
  • The spring behaves in a linearly elastic manner with stiffness k. If a concentrated load P is applied at B.

The spring force will be calculated as:-

F = kx

Deflection will be given by:-

x = \dfrac{PL^3}{3EI}

The spring force will be calculated by:-

F = \dfrac{KPL^3}{3EI}

The deflection of the beam will be given as:-

\rho = \dfrac{0.15KPL^3}{3EI}

Therefore the force in the spring will be F =\dfrac{KPl^3}{3EI}..The deflection of the beam will be \rho = 0.15(\dfrac{KPL^3}{3EI})

To know more about Cantilever beam follow

brainly.com/question/16791806

#SPJ1

7 0
2 years ago
simply supported beam is subjected to a linearly varying distributed load ( ) 0 q x x L 5 q with maximum intensity 0 q at B. The
Pavlova-9 [17]

Answer:

q₀ = 350,740.2885 N/m

Explanation:

Given

q(x)=\frac{x}{L} q_{0}

σ = 120 MPa = 120*10⁶ Pa

L=4 m\\w=200 mm=0.2m\\h=300 mm=0.3m\\q_{0}=? \\

We can see the pic shown in order to understand the question.

We apply

∑MB = 0  (Counterclockwise is the positive rotation direction)

⇒ - Av*L + (q₀*L/2)*(L/3) = 0

⇒ Av = q₀*L/6   (↑)

Then, we apply

v(x)=\int\limits^L_0 {q(x)} \, dx\\v(x)=-\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6} \\M(x)=\int\limits^L_0 {v(x)} \, dx=-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x

Then, we can get the maximum bending moment as follows

M'(x)=0\\ (-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x)'=0\\ -\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6}=0\\x^{2} =\frac{L^{2}}{3}\\  x=\sqrt{\frac{L^{2}}{3}} =\frac{L}{\sqrt{3} }=\frac{4}{\sqrt{3} }m

then we get  

M(\frac{4}{\sqrt{3} })=-\frac{q_{0}}{6*4} (\frac{4}{\sqrt{3} })^{3}+\frac{q_{0} *4}{6}(\frac{4}{\sqrt{3} })\\ M(\frac{4}{\sqrt{3} })=-\frac{8}{9\sqrt{3} } q_{0} +\frac{8}{3\sqrt{3} } q_{0}=\frac{16}{9\sqrt{3} } q_{0}m^{2}

We get the inertia as follows

I=\frac{w*h^{3} }{12} \\ I=\frac{0.2m*(0.3m)^{3} }{12}=4.5*10^{-4}m^{4}

We use the formula

σ = M*y/I

⇒ M = σ*I/y

where

y=\frac{h}{2} =\frac{0.3m}{2}=0.15m

If M = Mmax, we have

(\frac{16}{9\sqrt{3} }m^{2} ) q_{0}\leq \frac{120*10^{6}Pa*4.5*10^{-4}m^{4}   }{0.15m}\\ q_{0}\leq 350,740.2885\frac{N}{m}

8 0
3 years ago
What is code in Arduino to turn led on and off
11Alexandr11 [23.1K]

here's your answer..

4 0
3 years ago
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