X+y=16
the sum of the squares is
x^2+y^2=sum
solve for y in first equation
x+y=16
y=16-x
subsitiute that for y in other equation
x^2+(16-x)^2=sum
x^2+x^2-32x+256=sum
2x^2-32x+256=sum
take derivitive to find the minimum value (or just find the vertex because the parabola opens up)
derivitive is
4x-32=derivitve of sum
the max/min is where the derivitive equals 0
4x-32=0
4x=32
x=8
at x=8
so then y=16-8=8
the smallest value then is 8^2+8^2=64+64=128
Ratio of a leg to the hypotenuse = 1: sqrt2 where x = length of a leg, so
1 /sqrt2 = x / 8
x = 8 / sqrt2 = 8 sqrt2 / 2 = 4 sqrt2 or 5.66 to nearest hundredth answer
Answer:
Step-by-step explanation:
9. the set containing all objects or elements and of which all other sets are subsets.
10. complement --> the amount in only one of teh sets
intersection --> the amount in both sets
11. 14 or -14
13. 46 1/2
14. 2 31/120
15. 1100% profit
16. 1200 gm
17. cube root of 7?
18. 6000 per year
