y = x³ + 3x² - x - 3
0 = x³ + 3x² - x - 3
0 = x²(x) + x²(3) - 1(x) - 1(3)
0 = x²(x + 3) - 1(x + 3)
0 = (x² - 1)(x + 3)
0 = (x² + x - x - 1)(x + 3)
0 = (x(x) + x(1) - 1(x) - 1(1))(x + 3)
0 = (x(x + 1) - 1(x + 1))(x + 3)
0 = (x - 1)(x + 1)(x + 3)
0 = x - 1 or 0 = x + 1 or 0 = x + 3
+ 1 + 1 - 1 - 1 - 3 - 3
1 = x or -1 = x or -3 = x
Solution Set: {-3, -1, 1}
Answer:
There is no picture or graph to go with the question so I am afraid I will not be able to give you a specific answer.
To find out if a point (x, y) is on the graph of a line, we plug in the values into that equation and see if we get a true statement, such as 10 = 10. If we get something different, like 6 = 4, we know that the point is not on the line because it does not satisfy the equation. Plug in (-301, 601) into the equation of the line to see whether that point lies on it or not.
Step-by-step explanation:
Suppose the equation of the straight line that passes through E and F is this:
y = 7x + 2
We are to figure out whether or not the point (1, 10) lies on that line. In order to do this we would plug in (1, 10) into the equation, with 1 being x and 10 being y.
10 = 7(1) + 2 = 7 + 2 = 9
10 = 9 is a false statement. Therefore, the point (1, 10) does NOT lie on the line y = 7x + 2.
If you were to provide an image or graph that shows the equation of line AB then perhaps I would be able to answer your question with a specific answer.
Hello!
We will just do the math as shown below. We will use 4 7/7 to make the subtraction a bit easier.
4 7/7-4/7=4 3/7
Therefore, our answer is A) 4 3/7.
I hope this helps!
