Answer:I'm not sure but I can either divide or multiple
Step-by-step explanation:
Answer:
4:
a.) 49x^2-36y^2
b.) 9a^2-25b^2
c.) 1/25x^2-49
5: b.) -5x^2-4xy+4y^2
Step-by-step explanation:
So I'm going to assume that this question is asking for <u>non extraneous solutions</u>, or solutions that are found in the equation <em>and</em> are valid solutions when plugged back into the equation. So firstly, subtract 2 on both sides of the equation:
Next, square both sides:
Next, subtract x and add 2 to both sides of the equation:
Now we are going to be factoring by grouping to find the solution(s). Firstly, what two terms have a product of 6x^2 and a sum of -5x? That would be -3x and -2x. Replace -5x with -2x - 3x:
Next, factor x^2 - 2x and -3x + 6 separately. Make sure that they have the same quantity on the inside of the parentheses:
Now you can rewrite the equation as
Now, apply the Zero Product Property and solve for x as such:
Now, it may appear that the answer is C, however we need to plug the numbers back into the original equation to see if they are true as such:
Since both solutions hold true when x = 2 and x = 3, <u>your answer is C. x = 2 or x = 3.</u>
Answer:
Step-by-step explanation:
f(x) = x^2 -1
We need to find f^-1(x)
WE find inverse of f(x)
Replace f(x) with y
Replace x with y and y with x
Solve for y
add 1 on both sides ,
To remove square , take square root on both sides
Replace y with f^-1