Answer:
a = 1/2 (1 ±sqrt(47))
Step-by-step explanation:
a^2-a+12=0
We will complete the square
Subtract 12 from each side
a^2-a+12-12=0-12
a^2-a=-12
The coefficient of a = -1
-Divide by 2 and then square it
(-1/2) ^2 = 1/4
Add it to each side
a^2 -a +1/4=-12 +1/4
(a-1/2)^2 = -11 3/4
(a-1/2)^2= -47/4
Take the square root of each side
sqrt((a-1/2)^2) =sqrt(-47/4)
a-1/2 = ±i sqrt(1/4) sqrt(47)
a-1/2= ±i/2 sqrt(47)
Add 1/2 to each side
a-1/2+1/2 = 1/2± i/2 sqrt(47)
a = 1/2± i/2 sqrt(47)
a = 1/2 (1 ±sqrt(47))
Answer:
C
Step-by-step explanation:
The length of arc AC (L) is calculated as
L = circumference of circle × fraction of circle
= 2πr × 
= 2π × 12 × 
= 2π × 5
= 10π ≈ 31.42 → C
Ok so
since we h ave
3.5=1 day's morning
and
2.3=1 day's afternoon
we need to add them to find 1 day and multiply by 6 (number of days)
that answer is not listted so we make into anohter form
this wold be
6 times (3.5 +2.3)
if we distributed
a(b+c)=ab+ac
6(3.5+2.3)=6 times 3.5+6 times 2.3
this is D
<h3>
Answer: 53%</h3>
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Explanation:
The question asks "What percent of females participate in extracurricular activities?" This means we only focus on the second column. There are 36 women, and of this total, 19 are in extracurricular activities.
Dividing the two values leads to approximately 19/36 = 0.52777 which rounds to 0.53
Then we move the decimal to the right two spots to get 53%
Roughly 53% of the female students participate in extracurricular activities.
