Answer:
Step-by-step explanation:
y = 3x + 2
Slope m = 3
slope of the line perpendicular to this line = -3
Point (3,5)
y- y1 = m(x-x1)
y - 5 = -3(x-3)
y - 5 = -3*x-3 * -3
y - 5 = -3x + 9
y = -3x + 9 +5
y = -3x + 14
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
16:64
5:x (Cross Multiply)
16x = 64*5
16x = 320 (Then Divide both sides by 16 to isolate x)
x= 20