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2 is false.
1 is correct :D
I'm guessing there's an option D for assignment 2?
Yes, that's in the assessment 2 (continued) attachment
D is the correct ans for 2
Thomas is right :D
Notice how you get g(x)=0 for x=-3
Wait, I'm confused, how do I get the answer for number 2?
@thomas9, I think you would like to explain a bit more in order to help. I'll leave it to you as you find this mistake :D
Ok, first off: |x+3| is always positive, so it's not B or C, right?
From the graph in A you can see that g(x)=0 for x=3, but |3+3|=6 =/= 0, So it's not A either.
Which leaves D, furthermore from D we find g(x)=0 for x=-3, which is true: |-3+3|=0.
(g(x) is y in this case)
Don't we want a positive number and not 0?
0 is positive in my opinion, |x+3| is not negative if you like that better.
So is it clear?
Well yea, but how is the answer A not negative? They both have arrows going into the negative quarter
The vertical axis denotes |x+3|, if |x+3| were negative it would be in the lower half of the graph. With option A it just goes in the left half which means x becomes negative, but not |x+3|.
(I need the rest of the answers checked)
5 and 8 are wrong and I'm uncertain about 7 and 12. The rest is correct.
1. correct 2.incorrect 3. correct 4. correct 5. incorrect 6. Dunno 7. correct 8. incorrect 9. Dunno 10. correct 11. correct 12. correct
For 5, the one you've selected is a quadratic one, not a linear one. Linear equation should have a form like y=mx+c (c could be 0)
9. D) is correct since it's the only one where there isn't more than one value for a given x
For 8, you should check the y-int first, y-int is = 4 So, put x=0 into the equations, (b) would be rejected immediate;y and then check the x -int put y=0 into the equations, (a) and (d) will also be rejected. So the only option left is the answer
So five should be B?
and 8 is C?
8 correct 5 wrong, for direct variation, the y-int is 0, so c=0 in y=mx+c
It's y = mx + c but there isn't any equation with two numbers multiplying each other plus another number :/
don't forget that x+2 = 1x+2. Even if you don't see the "1", it's still there
So then f(x) = l x + 2 l is correct?
You mean question 5?
Yes question 5 sorry
But if what charron said is right, wouldn't that mean that x + 2 is correct since even if I don't see the one it's still there: x + 2 = 1x + 2?
yes, for question 5, since you're looking for a linear equation (owhich the form is y = mx+c, has you've stated), x+2 is definitely it. Also, please note that, if you had y =4, that would be a linear equation as well with m = 0 (y= 0x + c)
Ohh ok, thanks :)
Thank you everyone! I got 100%!!