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asambeis [7]
3 years ago
12

F. determine if (x + 3) is a factor of (x) = 2x3 + x2 –8x + 21 by using synthetic division. if so, find the other factors.

Mathematics
1 answer:
babymother [125]3 years ago
4 0
Since its x + 3 we use -3


-3| 2      1      -8      21           (these are the # without the x)
___________________

            -6     15     -21      
-------------------------------
    2      -5      7        0               
 
Drag the 2 down, multiple it by -3 to get -6 
1+(-6) = -5. -5 * -3 =15 . 15+(-8) = 7
7*-3 = -21. -21 +21 = 0 
Since last # is 0 we have a factor.


 factor is 2x^2 -5x -7 (we start with a power 1 less the original.)

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If you roll two fair dice repeatedly, what is the probability that you will get a sum of 4 before you get a sum of 5 ? (a) (b) (
AlexFokin [52]

Answer:

The probability that you will get a sum of 4 before you get a sum of 5 is \frac{1}{12}

Step-by-step explanation:

Given : If you roll two fair dice repeatedly.

To find : What is the probability that you will get a sum of 4 before you get a sum of 5 ?

Solution :

When two dice are rolled the outcomes are

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Total number of outcomes = 36

Favorable outcome get a sum of 4 before you get a sum of 5 is (1,3) ,(2,2) and (3,1) = 3

The probability that you will get a sum of 4 before you get a sum of 5 is

P=\frac{3}{36}

P=\frac{1}{12}

Therefore, The probability that you will get a sum of 4 before you get a sum of 5 is \frac{1}{12}

5 0
3 years ago
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