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Mnenie [13.5K]
3 years ago
13

10 in. 13 in. 16 in. What is the Surface area of the triangular prism

Mathematics
1 answer:
Marianna [84]3 years ago
3 0

Answer:

about 704 in 2

Step-by-step explanation:

surface area-2Ab+Pb*h

=2(13*8)+49.6(10)

=704

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Select all the correct answers.
Schach [20]

Answer:

....................

Step-by-step explanation:

5 0
2 years ago
In Paris, the sculpture Long-Term, created by Armand Fernandez, contains 60 cars embedded in 3.5 million pounds of concrete. How
enot [183]
You'll obviously have to convert "million pounds" to "tons."
Hint:  1 ton = 2000 pounds.  Create a conversion factor from this and then multiply "3.5 million pounds" by your conversion factor (with "pounds" in the denominator of this factor).
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
X + 3y =- 12<br> y+ 3 =- {(x – 3)
fenix001 [56]

Answer:

x = 6, y = -6

Step-by-step explanation:

solve for x and y

x + 3y =  -12

y + 3 = -x + 3

add

(x + 3y)  +  -x + 3  =  -12 + y + 3

3y + 3 = y - 9

2y + 3 = -9

2y = -12

y = - 6

x + 3(-6) = -12

x - 18 = -12

x = 6

8 0
3 years ago
Solve log2(x) – log2(3) = 2.
Lynna [10]

Answer:

x=9.64385618

Step-by-step explanation:

I used an app to solve it

3 0
3 years ago
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