If this is for E2020 the answer is
D)<span>No, his value of </span><span> should be positive because an even exponent indicates a positive value.
thats why it was wrong</span>
The solution to a system of (linear) equations is the point where the graphs intersect. Consider two parallel lines. By definition, two parallel lines never intersect each other, but all pairs of non-parallel lines will eventually intersect. That means they will also have a solution.
Let's consider what makes a line parallel to another line. It basically looks identical, having the same steepness (slope), but the graph is just shifted over. That is, a parallel line would have the same slope and a different y-intercept. For our equation

, or

in slope-intercept form, a parallel line will be of the form

.
That describes the form of a parallel line, which we do not want. Any other line, however, will give a solution to our system, so we merely want a line where the slope does not equal 2.
We can have any equation of the form

.
Answer:
3) (2,-9)
4) (0,-5)
5) (1,-8)
Step-by-step explanation:
3)
The vertex will occur between you x-intercepts.
You already found that happens at x=2.
To find the corresponding y-coordinate, replace x in
f(x)=(x+1)(x-5) with 2:
f(2)=(2+1)(2-5)
f(2)=(3)(-3)
f(2)=-9
So the vertex is (2,-9).
4)
The y-intercept is when x=0.
So in f(x)=(x+1)(x-5) replace x with 0:
f(0)=(0+1)(0-5)
f(0)=(1)(-5)
f(0)=-5
So the y-intercept is (0,-5).
5)
To find another point just plug in anything besides any x already used.
We preferably want to use a value of x that will keep us on their grid however far up,down,left, or right their grid goes out. So I'm going to choose something close to the vertex which is at x=2. Let's go with x=1.
So replace x in f(x)=(x+1)(x-5) with x=1:
f(1)=(1+1)(1-5)
f(1)=(2)(-4)
f(1)=-8
So another point to graph is (1,-8).
It should be 3 because you are going to be using the formation of 3√ 3•3•3
.
Have Monday equal to the variable X
Monday = X
Tuesday is 0.305 faster therefore
Tuesday = x+0.305