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bulgar [2K]
3 years ago
13

Use the functions f(x) = 4x − 5 and g(x) = 3x + 9 to complete the function operations listed below.

Mathematics
1 answer:
vlada-n [284]3 years ago
8 0
A.  (f + g)(x) = f(x) + g(x) = 4x - 5 + 3x + 9 = 7x + 4

B. f.g (x)   = f(x) * g(x) = (4x - 5)*(3x + 9) = 12x^2 + 36x -  15x - 45
                =  12x^2 + 21x - 45

C  Replace the x in f(x) by g(x):-
  =  4(3x + 9) - 5
  = 12x + 36 - 5
  = 12x  + 31
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All factors of x cubed minus 6x squared plus 11x minus 6
miv72 [106K]

Answer:

Step-by-step explanation:

x³-6x²+11x-6

put x=1

1³-6×1²+11×1-6=1-6=11-6=0

by synthetic division

1| 1 -6  11   -6

|     1  -5    6

|----------------

| 1  -5  6 |0

x²-5x+6=0

x²-2x-3x+6=0

x(x-2)-3(x-2)=0

(x-2)(x-3)=0

x=1

x-1=0

so x³-6x²+11x-6=(x-1)(x-2)(x-3)

6 0
3 years ago
(3n^2+13n^3+5n)-(7n+4n^3)
Fantom [35]

Answer: n*(3n+2)*(3n-1)

Step-by-step explanation:

4 0
3 years ago
In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experien
diamong [38]

Answer:

The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.

Step-by-step explanation:

Using Bayes' Theorem

P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|a)P(a)}

where

P(B|A) is probability of event B given event A

P(B|a) is probability of event B not given event A  

and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.

For this problem,

Let P(B1) = Probability of machine B1 = 0.3

P(B2) = Probability of machine B2 = 0.2

P(B3) = Probability of machine B3 = 0.5

Let P(D) = Probability of a defective product

P(N) = Probability of a Non-defective product

P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003

P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006

P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010

Likewise,

P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997

P(N|B2) be probability of a non-defective product produced by machine 2  = 1 - P(D|B2) = 1 - 0.006 = 0.994

P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990

For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)

P(B1|N) =\frac{P(N|B1)P(B1)}{P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)} = \frac{(0.297)(0.3)}{(0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)} = 0.1138

Hence the probability that a non-defective product is produced by machine B1 is 11.38%.

4 0
3 years ago
The Barkers started their trip with a full tank of gas and a total of 39,872 miles on their car. They stopped 4 hours later and
lara [203]

Answer:

27.5 miles per gallon

Step-by-step explanation:

3 0
2 years ago
AB = 5x -9; BC= -2x + 12
BaLLatris [955]

Answer:

X=-3

Step-by-step explanation:

5x - 9

-2x + 12

(+) (-)

7x = -21

X= -3

plz mark ice bear as Brainliest

3 0
2 years ago
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