(1 point) Find the limit, if it exists, or type DNE if it does not exist. A. lim(x, y)→(0,0)(x+5y)2x2+25y2=lim(x, y)→(0,0)(x+5y)2x2+25y2= B. lim(x, y)→(0,0)2x3+7y3x2+y2=lim(x, y)→(0,0)2x3+7y3x2+y2= (Hint for B: use polar coordinates, that is x=rcos(θ),y=rsin(θ)x=rcos⁡(θ),y=rsi n⁡(θ) )

A. DNE. The limit does not exist.

B. The limit is 0

Step-by-step explanation:

A. lim(x,y)→(0,0)[(x + 5y)/(2x² + 25y²)]

Along the x axis, y = 0, the limit becomes:

lim(x)→(0)[x/2x²]

= lim(x)→(0)[1/2x] = infinity

Along the y axis, x = 0, the limit becomes

lim(y)→(0)[5y/25y²]

= lim(y)→(0)[1/5y] = infinity

DNE

B. lim(x,y)→(0,0)[(2x³ + 7y³)/(x² + y²)

Putting x = rcosθ, and y = rsinθ, the limit becomes

lim(r)→(0)[(2r³cos³θ + 7r³sin³θ)/(r²cos²θ + r²sin²θ)]

Because cos²θ + sin²θ = 1, the limit becomes:

= lim(r)→(0)[r³(2cos³θ + 7sin³θ)/r²]

= lim(r)→(0)[r(2cos³θ + 7sin³θ)]

= 0

The limit is 0

step-by-step explanation:

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation: