(i) Yes. Simplify 
.
Now compute the limit by converting to polar coordinates.
This tells us
so we can define 
to make the function continuous at the origin.
Alternatively, we have
and
Now,
so by the squeeze theorem,
and approaches 1 as we approach the origin.
(ii) No. Expand the fraction.
and 
are undefined, so there is no way to make continuous at (0, 0).
(iii) No. Similarly,
is undefined when 
.
Ok so assume that if you have
xy=0 then x and y=0
-1/4(x^2-4x+3)>0
multiply both sides by -4 to clear fraction (-4/-4=1)
flip sign
x^2-4x+3<0
factor
(x-3)(x-1)<0
we know that (+) times (-)=(-) and
(-) times (-)=(+) so we don't want them to be both negative, we want different sign
we cannnot have 3 since it would be (0)(2)<0 which is false
we cannot have 1 either since (-2)(0)<0 is also false
lets see if the solution is in betwen 3 and 1 or outside of 3 and 1
ltry 2
2^3-4(2)+3<0
8-8+3<0
3<0
false
therefor it is outside ie
x>3 and x<1
6. I think it's b but I could be wrong hope I helped.
Answer:
(7,12),(-12,-7)
Step-by-step explanation:
x^2 + y^2 = 193
x - y = - 5
x = y - 5
(y - 5)^2 + y^2 = 193
y^2 - 10y + 25 + y^2 = 193
2y^2 - 10y + 25 = 193
2y^2 - 10y + 25 - 193 = 0
2y^2 - 10y - 168 = 0
2*(y^2 - 5y - 84) = 0 This factors into
2*( y - 12) (y + 7) = 0
y = 12
in which case x = 12 - 5 = 7
y = - 7
in which case x = -7 - 5 = - 12
