Answer:
2 sqrt(3) = n
k = 4 sqrt(3)
Step-by-step explanation:
We know that cos 30 = adjacent/ hypotenuse
cos 30 = 6 /k
Multiply each side by k
k cos 30 = 6k * k
k cos 30 = 6
Divide each side by cos 30
k cos 30/ cos 30 = 6/ cos 30
k = 6 / cos 30
k = 6/ (sqrt(3)/2)
k =12 / sqrt(3)
We cannot leave the sqrt in the denominator
k = 12 / sqrt(3) * sqrt(3)/sqrt(3)
k = 12 sqrt(3)/3
k = 4 sqrt(3)
We know tan 30 = opposite/adjacent
tan 30 = n/6
Multiply each side by 6
6 tan 30 = n
6 * (sqrt(3)/3) = n
6/3 * sqrt(3) = n
2 sqrt(3) = n
yes and.... please explain more.... I will make sure I edit my answer
A: 8*9= 72 or 72 because it is the lowest multiple of 8 and 9
B: euclidean theorem (35,63) (35,28) (7,28) (7,0) gcf= 7 or 7 because 7 is the biggest number that goes into both numbers.
C: 7(5+9) or 35 is a factored into 7 and 5 and 63 is factored into 7 and 9 and 9 is factored into 3 and 3
