How can (a+3)2=a2+6a+9 be derived from the polynomial identity (a−3)2=a2−6a+9? Add 12a to both sides of (a−3)2=a2−6a+9.
Substitute −a for a on both sides of (a−3)2=a2−6a+9.
Multiply both sides of (a−3)2=a2−6a+9 by −1.
Add 6 to both sides of (a−3)2=a2−6a+9.

the least common denominatore of 7 and 2 is 9

step-by-step explanation:

7+2=9

answer: 5

step-by-step explanation:

since, the range is the difference between the lowest and highest values of the given sample space.

and, here the sample space = { 3, 1, 0, 4, 1, 4, 2, 3, 0, 5, 0, 1, 1, 2, 0, 4, 1,4, 4 ,2 ,3 ,2 ,0 ,1 ,4 }

here, the highest value of the given sample = 5

and, the lowest value = 0

therefore, the range of the sample space = 5 - 0 = 5

answer: 75°

step-by-step explanation:

∠esk is the supplement of ∠edk, so is 100°. ∠esd = 1/2·arc ed, so is 25°.

since ∠esk = ∠esd + ∠dsk

you have 100° = 25° + ∠dsk . .

fill in known values 75° = ∠dsk . . . . .

subtract 25°

step-by-step explanation: