Answer:
Graph of a one-to-one function if f is a one-to-one function then no two points (x1,y1), (x2,y2) have the Same y-value. Therefore horizontal line cuts the graph of the equation y = f (x) more than omce. Horizontal line test: A graph passes the horizontal line test if each horizontal line cuts the graph at almost once.
Since there is a multiplier of 7 on the left, and both terms on the right are divisible by 7, I'd start by dividing the equation by 7
|8 - 3x| = 3x - 7
The absolute value function may or may not negate its argument. (The argument is negated if it is negative.) Hence this can resolve to two equations
-(8 -3x) = 3x -7
or
(8 -3x) = 3x -7
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Starting with the first of these, we can eliminate parentheses to get
... -8 +3x = 3x -7
Subtracting 3x from both sides gives the FALSE statement
... -8 = -7 . . . . FALSE
Thus, this version of the equation has No Solution.
Looking at the second of the equations above, we can add 3x to get
... 8 = 6x -7
Adding 7 and dividing by 6 gives
... (8 +7)/6 = x = 5/2
The solution is x = 5/2.
Answer:
3y² -2y -1 =0
Step-by-step explanation:
X= y²+1 .....(1)
3x - 2y = 4.....(2)
put x in (2) : 3(y²+1
)-2y =4
3y² +3 -2y -4 = 0
3y² -2y -1 =0
Answer:
y = (-2/3)x + 5/6
Step-by-step explanation:
The general structure of a line in slope-intercept form is:
y = mx + b
In this form, "m" represents the slope and "b" represents the y-intercept. You have been given the value of "m". To find the value of "b", you can plug the slope and the values from the point (2, -1/2) into the general equation.
m = -2/3
Point (2, -1/2):
x = 2 and y = -1/2
y = mx + b <----- Slope-intercept form
y = (-2/3)x + b <----- Insert -2/3 in "m"
-1/2 = (-2/3)(2) + b <----- Insert "x" and "y" values from point
-1/2 = -4/3 + b <----- Multiply -2/3 and 2
-3/6 = -8/6 + b <----- Give the fractions common denominators
5/6 = b <----- Add 8/6 to both sides
Thus, the equation of the line is:
y = (-2/3)x + 5/6
