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azamat
2 years ago
14

Which point is in the solution set of the equation 3x+2y>7

Mathematics
1 answer:
Mekhanik [1.2K]2 years ago
8 0

Answer:

it is a that is the answer

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Felicia is running a game booth at a carnival. One of the games requires participants to guess the weights, in grams, of a bag o
const2013 [10]
The weight of all 9 dimes is 20.43 grams
8 0
3 years ago
You are given 4 matrices M1, M2, M3, M4 and you are asked to determine the optimal schedule for the product M1 ×M2 × M3 ×M4 that
alexandr1967 [171]

Answer:

Step-by-step explanation:

first method is to try out all possible combinations and pick out the best one which has the minimum operations but that would be infeasible method if the no of matrices increases  

so the best method would be using the dynamic programming approach.

A1 = 100 x 50

A2 = 50 x 200

A3 = 200 x 50

A4 = 50 x 10

Table M can be filled using the following formula

Ai(m,n)

Aj(n,k)

M[i,j]=m*n*k

The matrix should be filled diagonally i.e., filled in this order

(1,1),(2,2)(3,3)(4,4)

(2,1)(3,2)(4,3)

(3,1)(4,2)

(4,1)

<u>                  Table M[i, j]                                             </u>

             1                      2                  3                    4

4    250000          200000        100000                0  

3      

750000        500000            0

2      1000000             0

1            

0

Table S can filled this way

Min(m[(Ai*Aj),(Ak)],m[(Ai)(Aj*Ak)])

The matrix which is divided to get the minimum calculation is selected.

Table S[i, j]

           1          2         3        

4

4          1           2         3

3          

1          2

2            1

1

After getting the S table the element which is present in (4,1) is key for dividing.

So the matrix multiplication chain will be (A1 (A2 * A3 * A4))

Now the element in (4,2) is 2 so it is the key for dividing the chain

So the matrix multiplication chain will be (A1 (A2 ( A3 * A4 )))

Min number of multiplications: 250000

Optimal multiplication order: (A1 (A2 ( A3 * A4 )))

to get these calculations perform automatically we can use java

code:

public class MatrixMult

{

public static int[][] m;

public static int[][] s;

public static void main(String[] args)

{

int[] p = getMatrixSizes(args);

int n = p.length-1;

if (n < 2 || n > 15)

{

System.out.println("Wrong input");

System.exit(0);

}

System.out.println("######Using a recursive non Dyn. Prog. method:");

int mm = RMC(p, 1, n);

System.out.println("Min number of multiplications: " + mm + "\n");

System.out.println("######Using bottom-top Dyn. Prog. method:");

MCO(p);

System.out.println("Table of m[i][j]:");

System.out.print("j\\i|");

for (int i=1; i<=n; i++)

System.out.printf("%5d ", i);

System.out.print("\n---+");

for (int i=1; i<=6*n-1; i++)

System.out.print("-");

System.out.println();

for (int j=n; j>=1; j--)

{

System.out.print(" " + j + " |");

for (int i=1; i<=j; i++)

System.out.printf("%5d ", m[i][j]);

System.out.println();

}

System.out.println("Min number of multiplications: " + m[1][n] + "\n");

System.out.println("Table of s[i][j]:");

System.out.print("j\\i|");

for (int i=1; i<=n; i++)

System.out.printf("%2d ", i);

System.out.print("\n---+");

for (int i=1; i<=3*n-1; i++)

System.out.print("-");

System.out.println();

for (int j=n; j>=2; j--)

{

System.out.print(" " + j + " |");

for (int i=1; i<=j-1; i++)

System.out.printf("%2d ", s[i][j]);

System.out.println();

}

System.out.print("Optimal multiplication order: ");

MCM(s, 1, n);

System.out.println("\n");

System.out.println("######Using top-bottom Dyn. Prog. method:");

mm = MMC(p);

System.out.println("Min number of multiplications: " + mm);

}

public static int RMC(int[] p, int i, int j)

{

if (i == j) return(0);

int m_ij = Integer.MAX_VALUE;

for (int k=i; k<j; k++)

{

int q = RMC(p, i, k) + RMC(p, k+1, j) + p[i-1]*p[k]*p[j];

if (q < m_ij)

m_ij = q;

}

return(m_ij);

}

public static void MCO(int[] p)

{

int n = p.length-1;     // # of matrices in the product

m    =    new    int[n+1][n+1];        //    create    and    automatically initialize array m

s = new int[n+1][n+1];

for (int l=2; l<=n; l++)

{

for (int i=1; i<=n-l+1; i++)

{

int j=i+l-1;

m[i][j] = Integer.MAX_VALUE;

for (int k=i; k<=j-1; k++)

{

int q = m[i][k] + m[k+1][j] + p[i-1]*p[k]*p[j];

if (q < m[i][j])

{

m[i][j] = q;

s[i][j] = k;

}

}

}

}

}

public static void MCM(int[][] s, int i, int j)

{

if (i == j) System.out.print("A_" + i);

else

{

System.out.print("(");

MCM(s, i, s[i][j]);

MCM(s, s[i][j]+1, j);

System.out.print(")");

}

}

public static int MMC(int[] p)

{

int n = p.length-1;

m = new int[n+1][n+1];

for (int i=0; i<=n; i++)

for (int j=i; j<=n; j++)

m[i][j] = Integer.MAX_VALUE;

return(LC(p, 1, n));

}

public static int LC(int[] p, int i, int j)

{

if (m[i][j] < Integer.MAX_VALUE) return(m[i][j]);

if (i == j) m[i][j] = 0;

else

{

for (int k=i; k<j; k++)

{

int   q   =   LC(p,   i,   k)   +   LC(p,   k+1,   j)   +   p[i-1]*p[k]*p[j];

if (q < m[i][j])

m[i][j] = q;

}

}

return(m[i][j]);

}

public static int[] getMatrixSizes(String[] ss)

{

int k = ss.length;

if (k == 0)

{

System.out.println("No        matrix        dimensions        entered");

System.exit(0);

}

int[] p = new int[k];

for (int i=0; i<k; i++)

{

try

{

p[i] = Integer.parseInt(ss[i]);

if (p[i] <= 0)

{

System.out.println("Illegal input number " + k);

System.exit(0);

}

}

catch(NumberFormatException e)

{

System.out.println("Illegal input token " + ss[i]);

System.exit(0);

}

}

return(p);

}

}

output:

7 0
3 years ago
Claire was out at a restaurant for dinner when the bill came. Her dinner came to $26. After adding in a tip, before tax, she pai
Kryger [21]

Answer:

0.0312

Step-by-step explanation:

You subtract 29.12 by 26 which equals 3.12

Then you divide by 100 which equals 0.0312

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2 years ago
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Please help ill mark brilliant​
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B is the answer to the question :)
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Which ratio is equivalent to 1:1
Lynna [10]

Answer:

the answer is 1

because 1 is not multiple in our self

6 0
3 years ago
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