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IrinaK [193]
2 years ago
6

Solve the equation -5(x-2)=-(x+6)

Mathematics
2 answers:
Tems11 [23]2 years ago
6 0

Steps to solve:

-5(x - 2) = -(x + 6)

~Simplify both sides

-5x + 10 = -x - 6

~Subtract 10 to both sides

-5x = -x - 16

~Add x to both sides

-4x = -16

~Divide -4 to both sides

x = 4

Best Of Luck!

AnnZ [28]2 years ago
3 0

Answer:

\huge{ \bold{ \huge{ \boxed{ \sf{x = 4}}}}}

Step-by-step explanation:

\sf{ - 5(x - 2) =  - (x + 6)}

Distribute -5 through the parentheses in left hand side

Similarly, distribute ( - ) sign through the parentheses in right hand side

\longrightarrow{ \sf{ - 5x + 10 =  - x  - 6}}

Move x to left hand hand side and change it's sign

Similarly, move 10 to right hand side and change it's sign

\longrightarrow{ \sf{ - 5x + x =  - 6 - 10}}

Collect like terms

\longrightarrow{ \sf{ - 4x =  - 6 - 10}}

The negative integers are always added but posses the negative ( - ) sign

\longrightarrow{ \sf{ - 4x =  - 16}}

Divide both sides by -4

\longrightarrow{ \sf{ \frac{ - 4x}{ - 4}  =  \frac{ - 16}{ - 4}}}

Calculate

\longrightarrow{ \sf{x = 4}}

Hope I helped!

Best regards! :D

You might be interested in
-
bulgar [2K]

Answer/Step-by-step explanation:

m<MAH = 90° (linear pair. It is also supplementary to m<MAK = 90°)

m<MAL + m<HAL = m<MAH

Which is:

(9x + 1) + (x + 9) = 90°

Solve for x

9x + 1 + x + 9 = 90

Collect like terms

9x + x + 1 + 9 = 90

10x + 10 = 90

Subtract both sides by 10

10x + 10 - 10 = 90 - 10

10x = 80

Divide both sides by 10

x = 8

Find m<MAL

m<MAL = 9x + 1

Plug in the value of x

m<MAL = 9(8) + 1 = 72 + 1 = 73°

Find m<HAL

m<HAL = x + 9

m<HAL = 8 + 9 = 17°

4 0
2 years ago
He ratio of the lengths of the corresponding sides of two rectangles is 8:3. the area of the larger rectangle is 320 ft2. what i
dusya [7]
The answer is:  [A]:  " 120 ft² " .
_______________________________________________________
Explanation:
_______________________________________________________
Set up a ratio as a fraction:
_______________________________________________________

       8/3 = 320/x ;
_______________________________________________________
Cross-multiply:
_____________________________________________________

8x = 3 * 320  ;

8x = 960 ;

Divide each side of the equation by "8" ;
 to isolate "x" on one side of the equation ; and to solve for "x" ;
______________________________________________________
   8x / 8  =  960 / 8 ;

to get:  x = 120 .

The answer is:  [A]:  " 120 ft² " .
______________________________________________________
3 0
3 years ago
What is the value of m (5m+100)
Elza [17]

o is the value of m i think

6 0
3 years ago
How much income of medicinal plants of Manipur ​
Ganezh [65]

Answer:

<em>The researchers documented 145 medicinal plants that the healers use for treating 59 ailments. They also found that the ethnic group used more than 40 species for treating more than one ailment.</em>

\huge\underline\red{Hope\:that\:helps}

3 0
2 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
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