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igomit [66]
3 years ago
10

Find the value of x² - x+1 when x=√3 -1, in the form a+b√3 where a and b are integers​

Mathematics
1 answer:
lidiya [134]3 years ago
5 0

Answer:

6-3\sqrt{3}

If you compare this to a+b\sqrt{3}, then a=6 and b=-3.

Step-by-step explanation:

We are given x=\sqrt{3}-1 and we are asked to find the value of x^2-x+1.

So we will need to find x^2.

x=\sqrt{3}-1

Square both sides:

x^2=(\sqrt{3}-1)^2

Expand using identity: (x-a)^2=x^2-2ax+a^2.

x^2=(\sqrt{3})^2-2(\sqrt{3})(1)+1^2

x^2=3-2\sqrt{3}+1

x^2=4-2\sqrt{3}

Let's go back to the full x^2-x+1.

x^2-x+1

(4-2\sqrt{3})-(\sqrt{3}-1)+1

4-2\sqrt{3}-\sqrt{3}+1+1

6-3\sqrt{3}

If you compare this to a+b\sqrt{3}, then a=6 and b=-3.

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(1-1/4)^5

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2 years ago
Find the equation of the line<br> That is perpendicular to y=-2x+1 and contains the point (8,2)
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Answer:

y=.5x-2

Step-by-step explanation:

If one line is perpendicular to another that means that their slopes are negative reciprocals

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so far we have

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A coin is weighted so that the probability of obtaining a head in a single toss is 0.3. If the coin is tossed 35 times, what is
Vlada [557]

Answer:

(1) The correct answer is option (a) 0.510

(2) The correct answer is option (b) = 0.9664

(3) The correct answer is option (a)  0.8665

Step-by-step explanation:

1. Find attached of question 1

2.

n= 850

p= 0.1400  

here mean of distribution(μ) =n*p

                                               =850* 0.1400

                                                = 119  

standard deviation σ = √(np(1-p))

                                    = √(850*0.1400(1-0.1400))

                                    = √102.34

                                       =10.1163  

for normal distribution z score =(X-μ)/σx    

therefore from normal approximation of binomial distribution and continuity correction:

probability = P(X>100.5)

                   = P(Z>-1.83)

                   = 1-P(Z<-1.83)

                   = 1-0.0336

                   = 0.9664

3.

n = 1000

p = 0.08

mean of distribution(μ) =np

                                       = 1000*0.08

                                        = 80

and standard deviation σ=√(np(1-p))

                                    = √(1000*0.08(1-0.08))

                                    = √736

                                    = 8.5790

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                   = P(Z>-1.11)

                    = 1-P(Z<-1.11)

                    = 1-0.1335

                    = 0.8665

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