The whole of the real numbers.
12
To solve the equation 2X+8=X+10 the first step is to get all instances of the variable 'X' on the same side of the 'equals' sign. To do this, we must first subtract 'X' from both sides of the equation: 2X+8-X=X+10-X which then becomes X+8=10. The next step is to leave the variable 'X' alone on it's side of the 'equals' by subtracting the numeral 8 from both sides of the equation: X+8-8=10-8 which then becomes X=2. To confirm that the solution is correct, we then substitute the solution we arrived at into the original equation in place of the variable 'X': 2*2+8=2+10 2*2=4 so: 4+8=2+10 4+8=12 and 2+10=12 so the final solution is: 12=12 Therefore the answer to the question "Is 2X+8 equal to X+10" is Yes, and the value of the variable 'X' is 2
No. The resulting equation has more solutions. For example, x = 2 has only one solution and that is x = 2.butx2= 4, the squared equation, has two solutions: x = +2 and x = -2No. The resulting equation has more solutions. For example, x = 2 has only one solution and that is x = 2.butx2= 4, the squared equation, has two solutions: x = +2 and x = -2No. The resulting equation has more solutions. For example, x = 2 has only one solution and that is x = 2.butx2= 4, the squared equation, has two solutions: x = +2 and x = -2No. The resulting equation has more solutions. For example, x = 2 has only one solution and that is x = 2.butx2= 4, the squared equation, has two solutions: x = +2 and x = -2
x^2 + y^2 = 100
x^2 + y^2 = 100 (x - 0 )^2 + ( y - 0)^2 = 10^2 This is now in the Pythagorean form. x^2 + y^2 = r^2 The centre in Cartesian coordinates is the displacement of (x,y). Since there is no displacement , the the centre is at (0,0) r^2 is the radius squared at 10^2 , then the radius has a length of '10'.
2
To plot three equations in MATLAB, you can use the fplot function for each equation or define them as functions. For example, you can create a script like this: f1 = @(x) x.^2; % First equation f2 = @(x) x + 2; % Second equation f3 = @(x) sin(x); % Third equation fplot(f1, [-10, 10], 'r'); hold on; % Plot first equation in red fplot(f2, [-10, 10], 'g'); % Plot second equation in green fplot(f3, [-10, 10], 'b'); % Plot third equation in blue hold off; grid on; legend('y=x^2', 'y=x+2', 'y=sin(x)'); This code sets the range for the x values, specifies colors for each plot, and adds a legend for clarity.
10
12
the solution is the answer to the equation. A solution is any value that makes the equation true. x + 2 = 10 has exactly one solution ....x=8 x + 2 > 10 has infinitely many solutions....x=9 or 10 or 11 or 12 or 13, etc
If the domain is the set of real numbers, so is the range.
10
If: 5x/10 = 2 then the value of x works out as 4
6 + 4 = 10; 5 x 2 = 10
10 x 2 = 20
x^2+1=10 x^2-9=0 (x-3)(x+3)=0 x=±3 the top line is the equation you are looking for
y ∈ ℜ