Answer:
Explanation:
1. To show that the equation −x=2 has a solution, we can solve for x by isolating it on one side of the equation.
Starting with −x=2, we can multiply both sides of the equation by -1 to get x = -2. Therefore, x = -2 is the solution to the equation −x=2.
2. To show that the equation 2xsin(x)+x^2=2 has a solution, we can analyze the equation to see if there is a possibility of a solution.
The equation involves trigonometric function sin(x), which oscillates between -1 and 1. We also have x^2 term, which can be positive or zero. Therefore, the sum of these terms (2xsin(x)+x^2) can take various values depending on the value of x.
If we evaluate the equation for different values of x, we will find that it indeed has solutions. For example, if we substitute x = 0, the equation becomes 0 = 2, which is not true. However, if we substitute x = π/2, the equation becomes 2 = 2, which is true. Therefore, x = π/2 is one of the solutions to the equation 2xsin(x)+x^2=2.
3. To show that the equation sin(x)+cos(x)=1 has a solution, we can use trigonometric identities to simplify the equation.
By applying the Pythagorean identity sin^2(x) + cos^2(x) = 1, we can rewrite the equation as sin(x) + cos(x) = sin^2(x) + cos^2(x).
Since sin^2(x) and cos^2(x) can both take values between 0 and 1, the sum of these terms can also take values between 0 and 2. Therefore, there are values of x for which sin(x) + cos(x) can equal 1.
For example, if we substitute x = π/4, the equation becomes sqrt(2)/2 + sqrt(2)/2 = 1, which is true. Therefore, x = π/4 is one of the solutions to the equation sin(x) + cos(x) = 1.
4. To show that the equation e^x+1+xe^x=100 has a solution, we can use algebraic methods to solve the equation.
Rearranging the equation, we get e^x + xe^x = 99.
Since e^x is a positive exponential function that increases rapidly, and x is a variable that can take various values, there is a possibility that the sum of these terms can equal 99.
To find an approximate solution, we can use numerical methods such as iteration or a graphing calculator. These methods can help us find an x value for which the equation is satisfied.
In conclusion, all four equations have solutions. The specific values of the solutions can be found through various methods depending on the equation, such as algebraic manipulation or numerical methods.