Answer:
286 students attended
148 non students attended
Step-by-step explanation:
Given
Solving (a): Number of students
Represent students with S and non students with N
So:
--- (1)
--- (2)
Make N the subject of formula in (1)
Substitute 434 - S for N in (2)
Open Bracket
Collect Like Terms
Divide through by -2
<em>Hence; 286 students attended</em>
Solving (b):
Recall that
<em>148 non students attended</em>
This graph will start at (.5, 0) as the vertex.
To find this, all you need to do if find the point where inside the absolute value sign is equal to 0. Since the graph can never have a negative value, this will be the lowest point.
2x - 1 = 0
2x = 1
x = .5
From there, you can plot each point by going up two and to the left one.
Examples of Points: (1.5, 2) and (2.5, 4)
You can also plot the other side of the graph by going up two and to the right one.
These points can be found using the slope, which is the number attached to x (2).
Example of Points (-.5, 2) and (-1.5, 4).
Answer:
Haha proofs are an interesting thing. Usually, nothing is to scale, which is why you can't measure anything. They are pretty annoying, but it helps to know why certain things are the way that they are and develop justification skills for higher level math.
Sorry to discourage you, but you're going to see "Justify" quite a lot in calculus and beyond which is basically a more informal version of a proof
you can never escape it tbh lol
Answer: False
=============================================
Explanation:
I'll use x in place of p.
The original equation 10x^2-5x = -8 becomes 10x^2-5x+8 = 0 after moving everything to one side.
Compare this to ax^2+bx+c = 0
We have
Plug those three values into the discriminant formula below
d = b^2 - 4ac
d = (-5)^2 - 4(10)(8)
d = 25 - 40*8
d = 25 - 320
d = -295
The discriminant is negative, which means we have no real solutions. If your teacher has covered complex or imaginary numbers, then you would say that the quadratic has 2 complex roots. If your teacher hasn't covered this topic yet, then you'd simply say "no real solutions".
Either way, this quadratic doesn't have exactly one solution. That only occurs when d = 0. Therefore, the original statement is false.
