Which equations have a leading coefficient of 3 and a constant term of –...
0 = 3x2 + 2x – 2
0 = –2 – 3x2 + 3
0 = –3x + 3x2 – 2
0 = 3x2 + x + 2
0 = –1x – 2 + 3x2
Explanation: The leading coefficient is the number that is to the left of the term with the largest exponent, which in this case is 2. The term 3x^2 is the leading term with the coefficient of 3. This is why the leading coefficient is 3 for choice A, choice C, and choice E. Choice B has a leading coefficient of -3 so we can rule that out. Choice D has a leading coefficient of 3, but the constant term is NOT -2. Instead the constant term is +2, so we can rule out choice D as well. The other choices that haven't been eliminated all have 3x^2 somewhere in them, as well as the constant term -2. The other x term isn't relevant to the restrictions placed in the instructions.
the correct option is 2.
the box plot represent the lower limit, first quartile, median, third quartile and upper limit of the data.
from the box plot of roadsters car it is noticed that,
from the box plot of bandits car it is noticed that,
using the above values we can say that the median of both cars is same but the mean of bandit car is less because all the values are same except upper limit and upper limit of bandit is less.
therefore option 2 is correct.
multiply each side by 6x to get rid of the fractions
6x^2 +4 = 6
subtract 4 from each side
6x^2 +4-4 = 6-4
divide by 6
6x^2/6 = 2/6
x^2 = 1/3
take the square root of each side
sqrt(x^2) = ±sqrt(1/3)