LifeIsADangerousGame
  • LifeIsADangerousGame
Hi everyone, I was wondering if someone could check my assessment for me. All of the answers are logged in, I just need someone to check them and tell me if they're right or wrong.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
welcome
LifeIsADangerousGame
  • LifeIsADangerousGame
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More answers

anonymous
  • anonymous
2 is false.
anonymous
  • anonymous
1 is correct :D
anonymous
  • anonymous
I'm guessing there's an option D for assignment 2?
LifeIsADangerousGame
  • LifeIsADangerousGame
Yes, that's in the assessment 2 (continued) attachment
anonymous
  • anonymous
D is the correct ans for 2
anonymous
  • anonymous
Thomas is right :D
anonymous
  • anonymous
Notice how you get g(x)=0 for x=-3
LifeIsADangerousGame
  • LifeIsADangerousGame
Wait, I'm confused, how do I get the answer for number 2?
anonymous
  • anonymous
@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
anonymous
  • anonymous
Ok, first off: |x+3| is always positive, so it's not B or C, right?
LifeIsADangerousGame
  • LifeIsADangerousGame
Right
anonymous
  • anonymous
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.
anonymous
  • anonymous
Which leaves D, furthermore from D we find g(x)=0 for x=-3, which is true: |-3+3|=0.
anonymous
  • anonymous
(g(x) is y in this case)
LifeIsADangerousGame
  • LifeIsADangerousGame
Don't we want a positive number and not 0?
anonymous
  • anonymous
0 is positive in my opinion, |x+3| is not negative if you like that better.
anonymous
  • anonymous
So is it clear?
LifeIsADangerousGame
  • LifeIsADangerousGame
Well yea, but how is the answer A not negative? They both have arrows going into the negative quarter
anonymous
  • anonymous
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|.
anonymous
  • anonymous
|dw:1333466168559:dw|
LifeIsADangerousGame
  • LifeIsADangerousGame
Ohhh
LifeIsADangerousGame
  • LifeIsADangerousGame
I see!
LifeIsADangerousGame
  • LifeIsADangerousGame
(I need the rest of the answers checked)
anonymous
  • anonymous
5 and 8 are wrong and I'm uncertain about 7 and 12. The rest is correct.
Callisto
  • Callisto
1. correct 2.incorrect 3. correct 4. correct 5. incorrect 6. Dunno 7. correct 8. incorrect 9. Dunno 10. correct 11. correct 12. correct
Callisto
  • Callisto
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)
anonymous
  • anonymous
9. D) is correct since it's the only one where there isn't more than one value for a given x
Callisto
  • Callisto
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
Callisto
  • Callisto
*immediately
LifeIsADangerousGame
  • LifeIsADangerousGame
So five should be B?
LifeIsADangerousGame
  • LifeIsADangerousGame
and 8 is C?
Callisto
  • Callisto
8 correct 5 wrong, for direct variation, the y-int is 0, so c=0 in y=mx+c
LifeIsADangerousGame
  • LifeIsADangerousGame
It's y = mx + c but there isn't any equation with two numbers multiplying each other plus another number :/
anonymous
  • anonymous
don't forget that x+2 = 1x+2. Even if you don't see the "1", it's still there
LifeIsADangerousGame
  • LifeIsADangerousGame
So then f(x) = l x + 2 l is correct?
Callisto
  • Callisto
You mean question 5?
LifeIsADangerousGame
  • LifeIsADangerousGame
Yes question 5 sorry
LifeIsADangerousGame
  • LifeIsADangerousGame
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?
anonymous
  • anonymous
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)
LifeIsADangerousGame
  • LifeIsADangerousGame
Ohh ok, thanks :)
LifeIsADangerousGame
  • LifeIsADangerousGame
Thank you everyone! I got 100%!!

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