7x+3+5x²
ax²+bx+c
5x²+7x+3
Hoped this helped!
Answer:
The correct options are: Interquartile ranges are not significantly impacted by outliers. Lower and upper quartiles are needed to find the interquartile range. The data values should be listed in order before trying to find the interquartile range. The option Subtract the lowest and highest values to find the interquartile range is incorrect because the difference between lowest and highest values will give us range. The option A small interquartile range means the data is spread far away from the median is incorrect because a small interquartile means data is nor spread far away from the median
To find the area multiply the length by the width.
The length and width of a square are the same.
Area = 15 x 15 = 225 square feet.
Answer:
A. (f + g)(1) = - 9
B. (f - g)(0) = -7
C. (fg)(3) = 0
Step-by-step explanation:
A. (f + g)(1) = f(1) + g(1) f(1) = 1^2 - 9 = 1 - 9 = - 8 g(1) = 1 - 2 = - 1
= -8 -1 = -9
B. (f - g)(0) = f(0) - g(0) f(0) = 0^2 - 9 = 0 - 9 = -9 g(0) = 0 - 2 = -2
= -9 + 2 = -7
C. (fg)(3) = f(3)(g(3) f(3) = 3^2 - 9 = 9 - 9 = 0 g(3) = 3 - 2 = 1
= 0(1) = 0
All are equal except A/D = C/B
I replaced the values with numbers
a=5 b=10 c=50 d=100
10/5=100/50 so it’s not b/a = d/c
5/10=50/100 so it’s not a/b = c/d
5/50 = 10/100 so it’s not a/c = b/d
5/100 is NOT 50/10 so it’s NOT a/d = c/b
