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2 years ago

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lskjdfghlskjfdhglkjkfhgldkjhgsldhglsjghlkjghlkjdfhglkdfjfjghldsfjghldjghdffgjhlsffghsldffghdfkj

1 answer:

vodka [1.7K]2 years ago

5 0

Answer:

(⌐■-■) (⌐■-■) (⌐■-■) (⌐■-■)

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As you push a toggle bolt into a wall, the

MakcuM [25]

Answer:

Explanation:
One the object is attached the toggle it's placed onto the bolt. The wings should be open in the direction of the bolt head. The toggle is than collapsed and the entire toggle bolt is inserted into the hole in the wall. Once the wings are fully through the opening they will spring open on the other side.
Hopt it's help thanks..

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2 years ago

Create a C language program that can be used to construct any arbitrary Deterministic Finite Automaton corresponding to the FDA

otez555 [7]

Answer:

see the explanation

Explanation:

/* C Program to construct Deterministic Finite Automaton */

#include <stdio.h>

#include <DFA.h>

#include <stdlib.h>

#include <math.h>

#include <string.h>

#include <stdbool.h>

struct node{

struct node *initialStateID0;

struct node *presentStateID1;

};

printf("Please enter the total number of states:");

scanf("%d",&count);

//To create the Deterministic Finite Automata

DFA* create_dfa DFA(){

q=(struct node *)malloc(sizeof(struct node)*count);

dfa->initialStateID = -1;

dfa->presentStateID = -1;

dfa->totalNumOfStates = 0;

return dfa;

}

//To make the next transition

void NextTransition(DFA* dfa, char c)

{

int tID;

for (tID = 0; tID < pPresentState->numOfTransitions; tID++){

if (pPresentState->transitions[tID].condition(c))

{

dfa->presentStateID = pPresentState->transitions[tID].toStateID;

return;

}

}

dfa->presentStateID = pPresentState->defaultToStateID;

}

//To Add the state to DFA by using number of states

void State_add (DFA* pDFA, DFAState* newState)

{

newState->ID = pDFA->numOfStates;

pDFA->states[pDFA->numOfStates] = newState;

pDFA->numOfStates++;

}

void transition_Add (DFA* dfa, int fromStateID, int(*condition)(char), int toStateID)

{

DFAState* state = dfa->states[fromStateID];

state->transitions[state->numOfTransitions].toStateID = toStateID;

state->numOfTransitions++;

}

void reset(DFA* dfa)

{

dfa->presentStateID = dfa->initialStateID;

}

5 0

3 years ago

The waffle slab is: a) the two-way concrete joist framing system. b) a one-way floor and roof framing system. c) the one-way con

GREYUIT [131]

Answer:

a) the two-way concrete joist framing system

Explanation:

A waffle slab is also known as ribbed slab, it is a slab which as waffle like appearance with holes beneath. It is adopted in construction projects that has long length, length more than 12m. The waffle slab is rigid, therefore it is used in building that needs minimal vibration.

4 0

3 years ago

Which of the following is not a function of the cooling system

Pie

Answer:

hola soy dora

Explanation:

5 0

2 years ago

. Using the Newton Raphson method, determine the uniform flow depth in a trapezoidal channel with a bottom width of 3.0 m and si

Over [174]

Answer:

y ≈ 2.5

Explanation:

Given data:

bottom width is 3 m

side slope is 1:2

discharge is 10 m^3/s

slope is 0.004

manning roughness coefficient is 0.015

manning equation is written as

$v =1/n R^{2/3} s^{1/2}$

where R is hydraulic radius

S = bed slope

$Q = Av =A 1/n R^{2/3} s^{1/2}$

$A = 1/2 \times (B+B+4y) \times y =(B+2y) y$

$R =\frac{A}{P}$

P is perimeter $= (B+2\sqrt{5} y)$

$R =\frac{(3+2y) y}{(3+2\sqrt{5} y)}$

$Q = (2+2y) y) \times 1/0.015 [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} 0.004^{1/2}$

solving for y $100 =(2+2y) y) \times (1/0.015) [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} \times 0.004^{1/2}$

solving for y value by using iteration method ,we get

y ≈ 2.5

5 0

3 years ago