Its 2x^2-4x, to check it add it to 3x^2-2x or
2x^(2)-4x+3x^(2)-2x
Answer: Answer A: Both classes have a range of 45
Step-by-step explanation:
above the two graphs we can see a number line ranging from 50-100
bellow that, we can see 2 box and whisker plots, one for second period and one for fourth period. the larger rectangular boxes show all the number results, the line in between them states the average result. Both lines have starting points at 55 and ending points at 100.
100-55=45 45 being the range. Since 2nd and 4th both share the same range answer A could possibly be correct
B is wrong because 2nd period has a median of 90
C is wrong because no periods have a median of 100
D is wrong because most of the students scored an 85+
conclusion: answer A is correct
Answer:
Given: circle
diameter = 10 cm => radius (R) = 5 cm
Find: measure of angle bounding sector = 11 π sq. cm.
Plan: determine what part of the circle’s total area equals the sector’s area.
Total Area of Circle A = π R^2 = π 5^2 = 25 π sq. cm.
Therefore: Sector Area = 11 π cm^2/25 π cm^2 = 11/25
Since the sector is 11/25 th of the circles area, the sector angle will measure 11/25 th of the circle’s circumference. They are proportional.
C = 2 π R = 2 π (5) = 10 π cm
Sector Arc = measure of sector angle = 11/25 (10 π) =
22π/5 radians
Answer: Sector Arc = 22π/5 Radians
Answer:
299 miles per hour
Step-by-step explanation:
![v=\frac{234}{\sqrt[3]{\frac{p}{w}}}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B234%7D%7B%5Csqrt%5B3%5D%7B%5Cfrac%7Bp%7D%7Bw%7D%7D%7D)
![v=\frac{234}{\sqrt[3]{\frac{1311}{2744}}}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B234%7D%7B%5Csqrt%5B3%5D%7B%5Cfrac%7B1311%7D%7B2744%7D%7D%7D)
Therefore, the velocity of the car at the end of a drag race is 299 miles per hour
