For -7pi/6 is an angle in second quadrant, then sine and cosecant must be positive; and cosine, secant, tangent and cotangent must me negative.
The reference angle is:
7pi/6-pi=7pi/6-6pi/6=(7pi-6pi)/6=pi/6
Then
sin(-7pi/6)=sin(pi/6)→sin(-7pi/6)=1/2
cos(-7pi/6)=-cos(pi/6)→cos(-7pi/6)=-sqrt(3)/2
csc(-7pi/6)=1/sin(-7pi/6)=1/(1/2)=1(2/1)=2/1→csc(-7pi/6)=2
sec(-7pi/6)=1/cos(-7pi/6)=1/(-sqrt(3)/2)=-1(2/sqrt(3))=-2/sqrt(3)→
sec(-7pi/6)=-[2/sqrt(3)]*sqrt(3)/sqrt(3)=-2sqrt(3)/[sqrt(3)]^2→
sec(-7pi/6)=-2sqrt(3)/3
tan(-7pi/6)=sin(-7pi/6)/cos(-7pi/6)=(1/2)/(-sqrt(3)/2)=-(1/2)*(2/sqrt(3))→
tan(-7pi/6)=-2/[2sqrt(3)]=-1/sqrt(3)=-[1/sqrt(3)]*[sqrt(3)/sqrt(3)]→
tan(-7pi/6)=-sqrt(3)/[sqrt(3)]^2→tan(-7pi/6)=-sqrt(3)/3
cot(-7pi/6)=cos(-7pi/6)/sin(-7pi/6)=[-sqrt(3)/2]/(1/2)=-sqrt(3)/2*(2/1)→
cot(-7pi/6)=-2sqrt(3)/2→cot(-7pi/6)=-sqrt(3)
Answers:
sin(-7pi/6) = 1/2
cos(-7pi/6) = - sqrt(3)/2
tan(-7pi/6) = - sqrt(3)/3
csc(-7pi/6) = 2
sec(-7pi/6) = - 2*sqrt(3)/2
cot(-7pi/6) = - sqrt(3)
(x - 3)^2 = 0
(x - 3)(x - 3) = 0
x^2 - 3x - 3x + 9 = 0
x^2 - 6x + 9 = 0 <===
Answer: the normal curve can be used as an approximation to the binomial probability considering the following condition: when the sample is large, in this case n=112
Step-by-step explanation:
for a binomial experiment to be approximated to normal distribution, the following conditions must be present:
i. sample size must be large, in this case sample size is 112
ii. the mean must be equal to np,where n is sample size and p is probability of success
iii. the standard deviation must be equal to npq,where q is the probability of failure
<u>Answer:</u>
Total Volume of composite figure = 635.2 cm³
<u>Steps:</u>
CV = h×3.14×(d/2)²
CV = 5×3.14(8/2)²
CV = 5×3.14(4)²
CV = 5×3.14×16
CV = 5×50.24
CV = 251.2 cm³
RV = h×w×l
RV = 4×8×12
RV = 4×96
RV = 384 cm³
TV = CV + RV
TV = 251.2 + 384
TV = 635.2 cm³
The answer is 2 y= 2 so 3•2=6 6-4=2
