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.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

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

I would attempt to eliminate the x^2 term and solve for y first then go back and solve for x

I did that but I got y=sqrt5

then I plugged it back into the first equation to solve for x and it came out as x=15.4

I'm not seeing how you obtained that

can you should your work so I can see how you first obtained y

how did you get sqrt(5)?

After adding the equations I get -y^2=-5

So y^2 +4y=5

you can factor y^2+4y-5

Help me with that?

what two numbers multiply to be -5 and add up to be 4?

think
-5 can be written as the following:
-5(1)
or
5(-1)
-5+1=?
5+(-1)=?

-4 or 4

so which pair of numbers satisfy my question?

"what two numbers multiply to be -5 and add up to be 4?"

oh 5 and -1

y^2+4y-5=0
(y+5)(y-1)=0

(x+5) (y-1) = 0

so now you just set both factor equal to 0
y+5=0 or y-1=0

y=-5 and or y=1

x=2

actually you might end up with 4 solutions since x has a square on it

yep there is one more solution to x^2=4
I assume that is where you got x=2 from

x also =-2

so we have (2,-5) and (-2,-5)
now we need to solve for x when y is 1

you can still use 2x^2+y^2=33

uh y=sqrt31

did you replace x with 1?

yeah

in other words we never said anything about x being 1

\[2x^2+y^2=33 \\ 2x^2+1^2=33 \\ 2x^2+1=33\]
solve for x

why are you replacing x with 1?
we never said anything about x being 1

Oh lol, I get x=4 after setting y=1

or also?

I assume you got that from x^2=16

but there is one more solution to that

x=-4

\[(2,-5); (-2,-5) ; (-4,1);(4,1)\]

sweet

okay now the question is which one of these is in quadrant 4 of the graph?

|dw:1433531008630:dw|

(2,-5)

|dw:1433531053701:dw|

yep sounds great

Yes:)

lol this again

|dw:1433531133632:dw|

Can you help me with something else?:)

I can try

Okay its pretty simple
what is the maximum number of solutions to this system
x^2+4y=64
and x+y=5

well one is a parabola and one is a line

what is the max number of times a line can intersect a parabola?

2 times

Awesome thanks

I have more but I think I am going to ask them in other questions so things dont get confusing

k

Wait so i was right in the other question? Dude convinced me otherwise. goddarnit.

wait no, here it was 2. i only got 1.

what other question?>