Answer:
A: 1 crackers = 95/4 cals
B: mass= 16/4
c: 15/4
C 28.84 just multiply 38.45 by .25 then you should get 9.6125 you subtract 38.45 by 9.6125 and get 28.84
Answer:
(i) (f - g)(x) = x² + 2·x + 1
(ii) (f + g)(x) = x² + 4·x + 3
(iii) (f·g)(x) = x³ + 4·x² + 5·x + 2
Step-by-step explanation:
The given functions are;
f(x) = x² + 3·x + 2
g(x) = x + 1
(i) (f - g)(x) = f(x) - g(x)
∴ (f - g)(x) = x² + 3·x + 2 - (x + 1) = x² + 3·x + 2 - x - 1 = x² + 2·x + 1
(f - g)(x) = x² + 2·x + 1
(ii) (f + g)(x) = f(x) + g(x)
∴ (f + g)(x) = x² + 3·x + 2 + (x + 1) = x² + 3·x + 2 + x + 1 = x² + 4·x + 3
(f + g)(x) = x² + 4·x + 3
(iii) (f·g)(x) = f(x) × g(x)
∴ (f·g)(x) = (x² + 3·x + 2) × (x + 1) = x³ + 3·x² + 2·x + x² + 3·x + 2 = x³ + 4·x² + 5·x + 2
(f·g)(x) = x³ + 4·x² + 5·x + 2
Answer:
A
Step-by-step explanation:
All pieces of data for seventh grade classes are larger than that of any kindergarten class on the table. This also proves with the mean provided.
The MAD, mean absolute deviation, helps you to identify the variation. The MAD for seventh grade was significantly larger than that of the kinder classes. The range for both sets of data is two times one another, further proving that 7th grade classes varies more.
C and D are irrelevant, and B says "varies less" which is the opposite of what's going on.
Hope that helps!
