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Svetradugi [14.3K]
3 years ago
7

Simplify (1+ V3) (2-3).

Mathematics
1 answer:
Harrizon [31]3 years ago
7 0
Answer: -1-3V
Explanation:
You might be interested in
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
Use the given facts about the functions to find the indicated limit
Alika [10]

Answer:

B. -12

Step-by-step explanation:

The given limit are;

\lim_{x \to -11} f(x)=-3 and \lim_{x \to -11} g(x)=4

We want to find;

\lim_{x \to -11} (fg)(x)=\lim_{x \to -11} f(x)\times \lim_{x \to -11} g(x)

We substitute the given limits to obtain;

\lim_{x \to -11} (fg)(x)=-3\times 4

\lim_{x \to -11} (fg)(x)=-12

3 0
3 years ago
2. A random sample of 30 households was selected as part of a study on electricity usage, and the number of kilowatt-hours (kWh)
Brrunno [24]

Answer:

Step-by-step explanation:

Given that n =30, x bar = 375 and sigma = 81

Normal distribution is assumed and population std dev is known

Hence z critical values can be used.

For 95% Z critical=1.96

Margin of error = 1.96(\frac{81}{\sqrt{30} } )=29

Confidence interval = 375±29

=(346,404)

B) 99% confidence

Margin of error = 2.59*Std error =38

Confidence interval = 375±38

=(337, 413)

C) For 90%

Margin of error = 20

Std error = 20/1.645 = 12.158

Sample size

n=(\frac{81}{12.158} )^2\\=44.38

Atleast 44 people should be sample size.

8 0
3 years ago
Select all that apply.
julia-pushkina [17]

Answer:

10/15 8/12 14/21

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Let line A be the graph of 5x + 8y = -9. Line B is perpendicular to line A and passes through the point (10,10). If line B is th
frozen [14]
Find slope of line A:
Move into slope-intercept form y = mx+b

<span>5x + 8y = -9
8y = -5x - 9
y = (-5/8)x - 9/8

The slope of line A is -5/8.

If </span><span>Line B is perpendicular to line A, then

slope Line B = negative reciprocal of slope Line A</span>
<span>slope Line B = 8/5

So like B has the equation

y = (8/5)x + b

If it passes through (10,10), we know that when x = 10, y = 10. Use those values to solve for b:
</span>
<span>y = (8/5)x + b
10 = (8/5)·10 + b</span>
<span>10 = (8)·2 + b
10 = 16 + b
b = -6

So line B has equation </span>
<span>y = (8/5)x - 6

m = 8/5 and b = -6

so

m + b = 8/5 - 6 = 8/5 - 30/5 = -22/5

So m+b = -22/5 or -4.4 in decimal form
</span>
6 0
3 years ago
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