4x^2+4x^2-20x+12y-2=0 standard form​

4x²+4y²-20x+12y-2=0

4x²-20x+4y²+12y = 2

4(x²-5x) + 4(y²+3y) = 2

4(x²-5x+25/4) + 4(y²+3y+9/4) = 2 + 4*25/4 + 4*9/4

4(x-5/2)² + 4(y+3/2)² = 36

(x-5/2)² + (y+3/2)² = 9

Step-by-step explanation:

hope this helped :)

y= -2/3x^2 + 5/3x + 1/6

Step-by-step explanation:

the dimensions are:

step-by-step explanation:

we want to find the dimensions of a rectangular box with volume of

we factor to obtain;

we rewrite the rightmost factor to obtain;

recall that:

we apply difference of two squares formula to obtain;

hence the dimensions are

the answer is c

step-by-step explanation: