anonymous
  • anonymous
can I please get some help for this Algebra question?
Mathematics
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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions.

anonymous
  • anonymous
can I please get some help for this Algebra question?
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
I don't know why I am so confused with this question! Can someone please point me in the right direction?
misty1212
  • misty1212
HI!!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

misty1212
  • misty1212
look at the graph what is \(f(-2)\)?
anonymous
  • anonymous
\[(f+g)(a)=f(a)+g(a)\]\[(f-g)(a)=f(a)-g(a)\]\[(fg)(a)=f(a) \times g(a)\]\[(\frac{f}{g})(a)=\frac{f(a)}{g(a)} \space ; \space g(a)\neq 0\] Use these formulae and find the values from the graph
anonymous
  • anonymous
so to find for your first part \[(f+g)(-2)=f(-2)+g(-2)\] you do it like that and the value of f(-2) and g(-2) may be obtained from the graph
anonymous
  • anonymous
Okay but I how do I find that?
anonymous
  • anonymous
F=-2,4? G=-2,0??
anonymous
  • anonymous
yep! what are the y values for these points
anonymous
  • anonymous
Think of your f(x) as y-values for the blue curve and g(x) as the y-values for the pink straight line so to find f(-2), we find the corresponding y value for the x value -2 on the blue curve and to find g(-2), we find the corresponding y value for the x value -2 on the pink straight line
anonymous
  • anonymous
Hmm..okay..Now what? :/ still kind of confused
anonymous
  • anonymous
you've found your \[f(-2)=4\]\[g(-2)=0\] Because -2,4 is a point lying on the blue curve and -2,0 is a point lying on the pink straight line
anonymous
  • anonymous
basically u look at the x value, look for what value of y for that x we have a point lying on the curve and the line
anonymous
  • anonymous
Ok do I add or subtract the numbers now? I think I remember learning that.
anonymous
  • anonymous
\[(f+g)(-2)=f(-2)+g(-2)\] add them up :)
anonymous
  • anonymous
-4 ?
anonymous
  • anonymous
|dw:1444071199043:dw|
anonymous
  • anonymous
Sorry got confused there. Now I get it...Lol. Ok for the next one, f(1)= 1 right?
anonymous
  • anonymous
yep!
anonymous
  • anonymous
and g(1)=3...so answer would be 4 also?
anonymous
  • anonymous
No! this time you are subtracting! \[(f-g)(1)=f(1)-g(1)\]
anonymous
  • anonymous
Ok I see that now lol So it would be 2
anonymous
  • anonymous
Nevermind, -2!
anonymous
  • anonymous
Good!
anonymous
  • anonymous
Remember, for the next one you have to multiply!
anonymous
  • anonymous
Yes I got 0 on that one..Now for D.
anonymous
  • anonymous
yep!
anonymous
  • anonymous
I got 1/3. Thanks SO much for your help. Have a great day!!
anonymous
  • anonymous
correct!!
anonymous
  • anonymous
An interesting thing to note here, if by chance your g(1) was 0, you would say that that your (f/g)(1) is not defined, as you can't divide by 0 \[(\frac{f}{g})(1)=\frac{f(1)}{g(1)}=\frac{f(1)}{0}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.