To solve a (2 \times 1) matrix, you typically perform operations such as addition, subtraction, or scalar multiplication. A (2 \times 1) matrix consists of two rows and one column, representing a vector in two-dimensional space. For example, if you have a matrix (\begin{pmatrix} a \ b \end{pmatrix}), you can add it to another (2 \times 1) matrix (\begin{pmatrix} c \ d \end{pmatrix}) by adding corresponding elements to get (\begin{pmatrix} a+c \ b+d \end{pmatrix}). If you're looking to solve an equation involving (2 \times 1) matrices, you would typically apply matrix operations or use systems of equations techniques based on the context.
To solve x + 1.4 = 2, subtract 1.4 from each side; x = 0.6
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First, solve for y.y=-8xif x=2, then y=-16.
To solve -2+x=7, you need to isolate the variable x by adding two to both sides of the equation. -2+x=7 -2+x+2=x 7+2=9 x=9
To solve the equation (6x - 2 = 10), first, add 2 to both sides to isolate the term with (x): (6x = 12). Then, divide both sides by 6 to solve for (x): (x = 2). Thus, the solution to the equation is (x = 2).
The equation to solve is given by. |-2 x + 2| -3 = -3 Add 3 to both sides of the equation and simplify. |-2 x + 2| = 0 |-2 x + 2| is equal to 0 if -2 x + 2 = 0. Solve for x to obtain. So, x = 1
That depends a great deal on the equation that x + 2 should be a part of. In order to solve for x, x + 2 needs to equal something. If x + 2 = 0, x = -2 If x + 2 = 4, x = 2
3x - 4 = -2 The answer is x=2/3
To solve x + 1.4 = 2, subtract 1.4 from each side; x = 0.6
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You cannot solve log x- 2 unless (i) log x - 2 is equal to some number or (ii) x is equal to some number.
First, solve for y.y=-8xif x=2, then y=-16.
To solve -2+x=7, you need to isolate the variable x by adding two to both sides of the equation. -2+x=7 -2+x+2=x 7+2=9 x=9
x=-2
The answer is x = 2.
If -x - 2 = 9, we can solve for x by first adding 2 to both sides: -x - 2 + 2 = 9 + 2 -x = 11. Next, multiply both sides by -1 to remove the negative and solve for x. -(-x) = -11 x = -11.
log9(x)=2 x=9^2 x=81