Its 35543562454645656466 and 3568899687698797997978 also 5679697-5679567-79567 two more 8439485949834595893458485 abd 84853945934853495834593458
Answer:
f[g(3)] = 100
g[f(5)] = 138
g{f[g(-4)]} = -982
Step-by-step explanation:
f(x) = 4x - 3
g(x) = 8x + 2
a. f[g(3)]
First find g(3) by putting x = 3 in g(x) function
g(3) = 8(3) + 2 = 24 +2 = 26
now put this g(3) = 26 as x in f(x)
f(g(3)) = f(26) = 4(26) - 4 = 104 - 4 = 100
b. g[f(5)]
First lets find f(5)
f(x) = 4x - 3
put x = 5 above
f(5) = 4(5) - 3 = 20 - 3 = 17
put this f(5) = 17 as x in g(x)
g(f(5)) = g(17) = 8(17) + 2 = 138
c.g{f[g(-4)]}
First lets solve the inner most function
g(-4) = 8(-4) +2 = -30
put g(-4) = -30 in f(x) to find f(g(-4))
f(g(-4)) = f(-30) = 4(-30) - 3 = -123
put f(g(-4)) = -123 as x in g(x) to find our complete result
g{f[g(-4)]} = g(-123) = 8(-123) + 2 = -982
I believe the answer would be d because 7+11 is not greater than 18
Answer:
the midpoint of the class with the greatest frequency 1500
Step-by-step explanation:
The illustration of the dataset given can be well represented in a table format as shown below.
Class Frequency Relative Frequency
$600 - $800 3 
$800 - $1000 7 
$1000 - $1200 11 
$1200 - $1400 22 
$1400 - $1600 40 
$1600 - $1800 24 
$1800 - $2000 9 
$2000 - $2200 4 
Total 120 1
Therefore, the midpoint of the class with the greatest frequency is between $1400 - $1600 since the frequency 40 happens to be the greatest frequency
The midpoint =
The midpoint =
= 1500
Answer:
1.Radious and diameter of a circle