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drek231 [11]
4 years ago
10

Which of the following is the complete list of roots for the polynomial function (x)= (x2+bx+3)(x+6x+13)?

Mathematics
1 answer:
shusha [124]4 years ago
4 0

Answer:

if i got you question right then its x=(-13/7) x=(-3/(2+b))

Step-by-step explanation:

to find the roots you need to make it equal to 0.

(x+6x+13)(2x+bx+3) =0

(7x+13)((2+b)x+3) =0

now there are 2 cases:

-> 7x+13=0 7x=(-13) x=(-13/7) solved

-> (2+b)x+3=0 (2+b)x=(-3) x=(-3/(2+b)) solved

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There is 13 questions that are worth 10 pints and 12 questions that are worth 12 points.

Step-by-step explanation:

8 0
3 years ago
PLEASE HELP!!!!!!!!!!!!!
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Is there an image or some information about the line?

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4 0
3 years ago
Algebra problem help
True [87]
To solve this problem you must apply the proccedure below:

 1. You have that one canned juice drink is 20% orange juice and another is 5% orange juice.

 2. You must make a system of equation, as below:

 x+y=15  (i)
 0.20x+0.05y=0.15x15   (ii)

 3. Now, you must find the value of x (the liters of the first canned needed) and y (the liters of the other canned needed):

 x+y=15  (i)
 x=15-y

 0.20x+0.05y=0.15x15   (ii)
 0.20(15-y)+0.05y=2.25
 y=5 liters

 x+y=15  (i)
 x=15-5
 x=10 liters

 4. Therefore, the asnwer is:

 10 liters of the canned juice drink that is 20% orange juice and 5 liters of the other canned juice drink that is 5% orange juice.

 
4 0
3 years ago
Find the length of the curve yequalsthree fifths x superscript 5 divided by 3 baseline minus three fourths x superscript 1 divid
Kazeer [188]
The answer is 333  so now it is equal
3 0
4 years ago
Based on past experience, 1 % of the telephone bills mailed to house-holds in Hong Kong are incorrect. If a sample of 10 bills i
nikdorinn [45]
<h2>Answer with explanation:</h2>

Given : The probability that telephone bills mailed to house-holds in Hong Kong are incorrect.=0.01

<h2>Binomial distribution :-</h2>

P(x)=^nC_xp^x(1-p)^{n-x}

If a sample of 10 bills is selected, then the probability that at least one bill will be incorrect :-

P(x\geq1)=1-P(0)\\\\=1-^{10}C_{0}(0.1)^0(0.9)^{10}\\\\=1-(0.9)^{10}=0.6513

Hence, the probability that at least one bill will be incorrect =0.6513

<h2>Poisson distribution: </h2>

P(x;\mu)=\dfrac{e^{-\mu}\mu^x}{x!}

Mean : \mu=np=10\times0.1=1

Then , If a sample of 10 bills is selected, then the probability that at least one bill will be incorrect :-

P(x\geq1)=1-P(0)\\\\=1-\dfrac{e^{-1}1^0}{0!}\\\\=1-0.3678=0.6321

Hence, the probability that at least one bill will be incorrect =0.6321

3 0
3 years ago
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