Answer and Step-by-step explanation:
a.)x^2 and y^2 are always greater than zero.(assuming real x and y).
=>x^2<=x^2+y^2
=>|x|(x^2)<=|x|(x^2+y^2)
=>|x^3|<=|x|(x^2+y^2)
b. similarly, |y^3|<=|y|(x^2+y^2)
=>|f(x,y)|=|x^3+y^3|/(x^2+y^2)=|x|+|y|
c. let h is very small number (close to zero but positive)
lim (x,y)-> (0,0) f(x,y)=(x^3+y^3)/(x^2+y^2)
putting x=h and y=h and approaching h->0
lim h->0 f(h)=2h=0
There is another explanation attached below
First, let's find the x and y intercepts
In the first equation
y - 4x = -1
Put x =0
y= - 1
(0, -1)
Put y=0 and the n solve for x
0 - 4x = -1
-4x = -1
x=0.25
(0.25 , 0)
The points for the first equation is (0, -1 ) and (0.25, 0)
Next is to find the intercts for the second equation
y + x = 4
put x=0
y = 4
(0, 4)
Put y =0
0 + x = 4
x = 4
( 4, 0)
The points for the second equation are;
(0, 4) and (4, 0)
Below is the graph
Answer:
3y+4
Step-by-step explanation:
9y-6y=3y