There is only one solution set. Depending on the inequalities, the set can be empty, have a finite number of solutions, or have an infinite number of solutions. In all cases, there is only one solution set.

Negative times negative equals positve, so -a*-b=ab (positive ab)

yes, it is both symmetric as well as skew symmetric

/* multiplication of a 3*3 matrix*/

#include<stdio.h>

main()

{

int a[3][3],b[3][3],c[3][3];

int i,j,k;

printf("enter the elements in A matrix:\n");

for(i=0;i<=2;i++)

{

for(j=0;j<=2;j++)

{

scanf("%d",&a[i][j]);

}

}

printf("enter b matrix:\n");

for(i=0;i<=2;i++)

{

for(j=0;j<=2;j++)

{

scanf("%d",&b[i][j]);

}

}

for(i=0;i<=2;i++)

{

printf("\n");

for(j=0;j<=2;j++)

{

c[i][j]=0;

for(k=0;k<=2;k++)

{

c[i][j] = c[i][j]+a[i][k] * b[k][j];

}

}

}

printf("multiplication matrix is:\n");

for(i=0;i<=2;i++)

{

for(j=0;j<=2;j++)

{

printf("%d\t",c[i][j]);

}

printf("\n");

}

}

Grams measures mass. Centimeters measures length.

When they have the same arrangement of numbers.

30-3 5+2 6-1 1+2 3-1 4+10

- Yes

6-0 9-4 1+2 2•2 10-7 5+8

2-1 4+2 1-3

- No

6-2 9+1 4+2

When there is no ordered n-tuple (pair, triplet, quartet - depending on the number of dimensions) that satisfies both equations at the same time. In graph form, it is when the straight lines representing the two equations are non-intersecting.

In 2 dimensions non-intersection implies that the lines are parallel.

It is a quadratic expession and when factored is: (x-6)(x+7).

A system of equations is a set of equations with more than one variable dealing with the same material. If there are 2 variables, then the system must have 2 equations before it can be solved. 3 variables need 3 equations, etc.

It is 1.

Yes.

If you were to plot y=23 on a graph, you'd have a straight horizontal line where y=23 (because no matter what the value of x is on the graph, y is always 23).

As the line that is plotted is straight, the equation is considered linear.

-648

It reminds me nothing in particular that I can remember

A rate of change of 60 miles per hour.

Think of it in terms of a car. The car is going to go a father distance if it is going 60 miles per hour as opposed to 1 mile per hour.

Normally, you do not choose them: you calculate them.

9 or 27

If the lines cross then there is one solution. If they are on top of each other then there are infinite solutions. If they are parallel then there are no solutions.

Example:

- 454 g of orange jam contain 83 g sugar

454 g jam---------------83 g sugar
100 g jam----------------x g sugar

x= 100x83/454= 18,18 g sugar (or 18 % sugar)

It works out that x = 0 and y = 3

you probably can, apart from the brackets and curvy x

Let M be the subset of 2x2 matrices A such that det(A)=0 and tr(A'A)=4 (I shall use ' to denote transpose).

Recall that one of the three definitions of a k-dimensional manifold is:

M c R^n is a k-dimensional manifold if for any p in M there is a neighborhood W c R^n of p and a smooth function F:W-->R^n-k so that F^-1(0) = M intersection W and rank(DF(x))=n-k for every x in M intersection W.

In short, what we need to do is find a function F so that the inverse image of the zero vector under F gives M and the rank of the derivative of F is equal to the dimension of the codomain of F.

Let an arbitrary 2x2 matrix be written as:

[a c]

[b d]

Then the two constraints that define M are

1) ad-bc=0

2) a^2+b^2+c^2+d^2=norm(a,b,c,d)^2=4

Define F:R^4-->R^2 by F(a, b, c, d)=(ad-bc, norm(a,b,c,d)^2-4). Then clearly F^-1(0,0)=M. Furthermore, F is smooth on M because the multiplication and addition of smooth functions (a, b, c, d) is also smooth. (Note that we have taken the neighborhood W to be some superset of M. This guaranteed to exist because if we interpret the set of 2x2 matrices as R^4, every point in M has norm 2, so any ball centered at the origin with length greater than 2 will contain M).

All that remains to be done is to check that rank(DF(x))=2 for every x in M. Observe that [DF]= [d -c -b a]

[2a 2b 2c 2d]

We shall now argue by contradiction. Suppose rank(DF) did not equal 2 for every x in M. Then we know that the two rows are linearly dependent i.e.

h[d -c -b a] + k[2a 2b 2c 2d] = 0 and h and k are not both 0.

Suppose h is 0. Then we have 2ka = 2kb = 2kc = 2kd = 0, and since k cannot also be 0, this implies that a=b=c=d=0, therefore norm(a,b,c,d)^2=0. But, (a,b,c,d) must be in M, so this is a contradiction. Hence h cannot be 0.

Now suppose h is nonzero. Then we can divide it out and there exists a, b, c, d and a constant k so that

[d -c -b a] + k[2a 2b 2c 2d] = 0 i.e. we have:

d + 2ka = -c + 2kb = -b + 2kc = a + 2kd = 0. From this we can substitute to obtain:

a(1 - 4k^2) = b(4k^2 - 1) = c(4k^2 - 1) = d(1 - 4k^2) = 0 and hence

a^2(1 - 4k^2)^2 = b^2(4k^2 - 1)^2 = c^2(4k^2 - 1)^2 = d^2(1 - 4k^2)^2 = 0.

Note that (1 - 4k^2)^2 = (4k^2 - 1)^2. Now, adding the four above expressions together we get:

(a^2 + b^2 + c^2 + d^2)(1 - 4k^2)^2 = norm(a,b,c,d)^2(1 - 4k^2)^2 = 0. But, since we require (a,b,c,d) to be in M, this reduces to 4(1 - 4k^2)^2 = 0. This implies that

1 - 4k^2 = 0, and hence k = +/-(1/2).

Now, if k=+1/2, then we have a + d = b - c = 0, therefore d = -a and b = c. Since we have det(A) = 0 as one of our constraints on M, this implies that ad - bc =

-(a^2) - (b^2) = -(a^2 + b^2) = 0, which implies a = b = 0, by the property of norms. But, if a = b = 0, then (a,b,c,d) = (0,0,0,0) and hence norm(a,b,c,d)^2 = 0, which is a contradiction.

We can argue analogously for the case where k = -1/2. Hence, assuming that rank(DF) is not 2 for some (a,b,c,d) in M leads to a contradiction, so we conclude that rank(DF)=2 for all x in M.

Finally, from this result, we conclude that since F:R^4-->R^2=R^(4-2), M must be a 2-dimensional manifold.

A 7*2 matrix (not matrice) and a 2*6 matrix, if multiplied together, will from a 7*6 matrix.

If you mean: 8x2-64x+120 = 0 then x = 3 or x = 5

Complex vocal tics involve meaningful words, phrases or sentences