Assume something (e.g. equations) using k then prove k+1 using k.
To find 100 division equations that equal 100, you can start with the basic equation 100 ÷ 1 = 100. Then, you can divide multiples of 100 by numbers ranging from 1 to 100, such as 200 ÷ 2 = 100, 300 ÷ 3 = 100, and so on. You can also create equations with decimals, such as 50 ÷ 0.5 = 100. Additionally, you can use negative numbers, such as -100 ÷ -1 = 100, to come up with a variety of division equations that equal 100.
There is no quadratic equation that is 'linear'. There are linear equations and quadratic equations. Linear equations are equations in which the degree of the variable is 1, and quadratic equations are those equations in which the degree of the variable is 2.
Linear Equations are equations with variable with power 1 for eg: 5x + 7 = 0 Simultaneous Equations are two equations with more than one variable so that solving them simultaneously
To solve for 2 unknown variables you need 2 independent equations. You have only 1.
just create a mathematical equation using 1964 such as 1x1964 2x1964 3x1964 4x1964 or 1964/1 1964/2 etc
100
((9/9)-(9/9)+(9/9)) = 1
If you mean: x2+2x+1 = 0 then it is a quadratiic equations whose solutions are equal because x = -1 and x = -1
Assume something (e.g. equations) using k then prove k+1 using k.
All of them, but it does depend on what operators are allowed.
100-1 101-2 102-3
(2+3)(5)=(4)(6)+1 Is this what you mean?
3(5x-2y)=18 5/2x-y=-1
1= 3/3 2=(3+3)/3 3= 3+3-3 4=(3+3+3+3)/3 5=(3+3+3+3+3)/3 6=(3+3+3+3+3+3)/3 and so on
If you mean: 6x-3y = -33 and 2x+y = -1 Then solving the simultaneous equations by substitution: x = -3 and y = 5
Solving equations in two unknowns requires two independent equations. Since you have only one equation there is no solution. .