You have not shared the graphs from which you have to choose your answer.
Why not graph y=|x-4| yourself, to help you choose the correct answer?
First, graph y=|x|. It looks like a "v" opening up, with the vertex at (0,0).
Next, move the whole "v" graph 4 units to the right.
Compare your result to the given answer choices.
Item | Frequency
0 | 2
1 | 5
2 | 4
3 | 3
5 | 1
Hope I helped!
Answer:
No because they say we all need to talk as a community
Step-by-step explanation:
Answer:
This can be done in a total of 10 ways
Step-by-step explanation:
This is a selection problem.
What we are trying to do is to properly select 3 week long camps from the total 5. we are looking for the number of ways in which we can do this.
Now, since this is a selection problem, the proper mathematical term and approach to use is the COMBINATION
Mathematically, having r items to pick from a total n, the number of ways to do this is nCr ways
which is mathematically equivalent to;
n!/(n-r)!r!
now applying this to the problem at hand, what we have is 5C3
= 5!/(5-3)!3! = 5!/2!3! = (5 *4)/2 = 20/2 = 10 ways
Answer:
(- 2, 15 ) and (2, 7 )
Step-by-step explanation:
Given the 2 equations
y = 3x² - 2x - 1 → (1)
2x + y = 11 → (2) ← subtract 2x from both sides
y = 11 - 2x → (3)
Substitute y = 3x² - 2x - 1 into (3)
3x² - 2x - 1 = 11 - 2x ( subtract 11 - 2x from both sides )
3x² - 12 = 0 ( add 12 to both sides )
3x² = 12 ( divide both sides by 3 )
x² = 4 ( take the square root of both sides )
x = ± 
= ± 2
Substitute these values into (3) for corresponding values of y
x = - 2 : y = 11 - 2(- 2) = 11 + 4 = 15 ⇒ (- 2, 15 )
x = 2 : y = 11 - 2(2) = 11 - 4 = 7 ⇒ (2, 7 )
