Which statement is correct?
(2.06 times 10 Superscript negative 2 Baseline) (1.88 times 10 Superscript negative 1 Baseline) less-than StartFraction 7.69 times 10 Superscript negative 2 Baseline Over 2.3 times 10 Superscript negative 5 Baseline EndFraction
(2.06 times 10 Superscript negative 2 Baseline) (1.88 times 10 Superscript negative 1 Baseline) greater-than-or-equal-to StartFraction 7.69 times 10 Superscript negative 2 Baseline Over 2.3 times 10 Superscript negative 5 Baseline EndFraction
(2.06 times 10 Superscript negative 2 Baseline) (1.88 times 10 Superscript negative 1 Baseline) greater-than StartFraction 7.69 times 10 Superscript negative 2 Baseline Over 2.3 times 10 Superscript negative 5 Baseline EndFraction
(2.06 times 10 Superscript negative 2 Baseline) (1.88 times 10 Superscript negative 1 Baseline) = StartFraction 7.69 times 10 Superscript negative 2 Baseline Over 2.3 times 10 Superscript negative 5 Baseline EndFraction

C

Step-by-step explanation:

(2.06 times 10 Superscript negative 2 Baseline) (1.88 times 10 Superscript negative 1 Baseline) is rounded to 3.9

StartFraction 7.69 times 10 Superscript negative 2 Baseline Over 2.3 times 10 Superscript negative 5 Baseline EndFraction  is rounded to 3.3

3.9 > 3.3

Option C: (2.06 × 10^-2)(1.88 × 10^-1) > 7.69 × 10^-2 / 2.3 × 10^-5

--

2.06 × 10^-2 = 206

1.88 × 10^-1 = 18.8

206 × 18.8 = 3,872.8

-

7.69 × 10^-2 = 769

2.3 × 10^-5  = 230,000

769 ÷ 230000 = ‭0.00334347826086956521739130434783‬

-

3,872.80 > 0.00334347826086956521739130434783‬

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2

b) k’(6, 4); l’(3, 3)

step-by-step explanation:

a 270° clockwise rotation maps every point (x, y) to (y, -x). (it negates the x-coordinate and then switches it and the y-coordinate.)

this means the endpoints of the side congruent to kl, k' and l', would be mapped as:

k'(-4, 6)→(6, 4)

l'(-3, 3)→(3, 3)