Answer:
whats the wuestion tho
Step-by-step explanation:
Answer:
6x - 1
Step-by-step explanation:
Side 1: 5x - 1
Side 2: 4x + 4
Side 3: Unknown, y
Perimeter: 15x + 2
Perimeter = Side 1 + Side 2 + Side 3
15x + 2 = 5x - 1 + 4x + 4 + y
15x + 2 = 9x + 3 + y
6x - 1 = y
y = 6x - 1
Side 3 = 6x - 1
Answer:
No solution
Step-by-step explanation:
Use substitution
2x+1=2x-1
Subtract 2x on both sides
1= -1
No solution
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>
First, start with h(x) and plug in 3
h(3)=3^(2)+4
And you get 13. Next, plug the 13 into the g(x):
g(13)=2(13)
And you get 26.