Answer:
Step-by-step explanation:
Let the solution to
2x^2 + x -1 =0
x^2+ (1/2)x -(1/2)
are a and b
Hence a + b = -(1/2) ( minus the coefficient of x )
ab = -1/2 (the constant)
A. We want to have an equation where the roots are a +5 and b+5.
Therefore the sum of the roots is (a+5) + (b+5) = a+ b +10 =(-1/2) + 10 =19/2.
The product is (a+5)(b+5) =ab + 5(a+b) + 25 = (-1/2) + 5(-1/2) + 25 = 22.
So the equation is
x^2-(19/2)x + 22 =0
2x^2-19x + 44 =0
B. We want the roots to be 3a and 3b.
Hence (3a) + (3b) = 3(a+b) = 3(-1/2) =-3/2 and
(3a)(3b) = 9(ab) =9(-1/2)=-9/2.
So the equation is
x^2 +(3/2) x -9/2 = 0
2x^2 + 3x -9 =0.
Each student will get a boot up the anus
Parallel to the y axis...that would mean that it is a vertical line.
The equation would be : x = 4 and it would have an undefined slope
One input can map out multiple outputs
Answer:

Step-by-step explanation:

Let's substitute the first equation into the second one.

- Multiply by
and bring all the terms to one side.
I'll use my graphical calculator to solve this but you could factor:
(All rounded to 2 decimal places)
Now we can substitute these
values into an equation for
.

