a+b+c=0
[(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc]
[a^2+b^2+c^2+2ab+2ac+2bc=0]
[a^2+b^2+c^2=-(2ab+2ac+2bc)]
[a^2+b^2+c^2=-2(ab+ac+bc)] (i)
also
[a=-b-c]
[a^2=-ab-ac] (ii)
[-c=a+b]
[-bc=ab+b^2] (iii)
adding (ii) and (iii) ,we have
[a^2-bc=b^2-ac] (iv)
devide (i) by (iv)
[(a^2+b^2+c^2)/(a^2-bc)=(-2(ab+bc+ca))/(b^2-ac)]
D. 5x = 45 because you are trying to find x, the number of tickets that must be sold to earn $45.
<span>The median would be preferred over the mean in such scenarios because the median will lessen the impact of the outliers that fall within the "tail" of the skew. Therefore, if a curve is normally distributed, that is to say that data is normally distributed, there will be two tails, each with approximately equal proportions of outliers. Outliers in this case being more extreme numbers, and are based on your determination depending on how you are using the data. If data is skewed there is one tail, and therefore it may be an inaccurate measure of central tendency if you use the mean of the numbers. Thinking of this visually. In positively skewed data where there is a "tail" towards the right and a "peak" towards the left, the median will be placed more in the "peak", whereas the mean will be placed more towards the "tail", making it a poorer measure of central tendency, or the center of the data.</span>
<span>ΔABC is not a right triangle.</span>
V=s³
SA=6s²
sub 2 for s
V=2³=8 cubic inches
SA=6(2²)=6(4)=24 square inches
