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3 years ago

12

WHIHIHIHI ) Question 6 2 pts Given the first term and the recursive formula for a sequence generate the first five terms. f(1) =

13 f(n) = f(n-1) - 3

1 answer:

maks197457 [2]3 years ago

5 0

Answer:

6wusyhdodnbdhsggegehd

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Jasmine said that if you are rounding 234 to 1sf, then you do not have to look at the hundreds in the question. Do you agree wit

Montano1993 [528]

也是哦去过热的with赫然 because舍is right

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3 years ago

1. 8y^2-4y+1 divided by 2y-1

____ [38]

Q1. The answer is $4y+ \frac{1}{2y-1}$

$\frac{8y^{2}-4y+1 }{2y-1} = \frac{4y*2y-4y+1}{2y-1} = \frac{4y(2y-1)+1}{2y-1} = \frac{4y(2y-1)}{2y-1}+ \frac{1}{2y-1} =4y+ \frac{1}{2y-1}$

Q2. The answer is $2a+3+ \frac{6}{a-1}$

$\frac{2 a^{2}+a+3 }{a-1} = \frac{a*2a-2a+3a+3}{a-1} = \frac{2a(a-1)+3a+3}{a-1}= \frac{2a(a-1)}{a-1}+ \frac{3a+3}{a-1} = \\ \\ =2a+ \frac{3a+3}{a-1}=2a+ \frac{3a-3+3+3}{a-1}=2a+ \frac{3(a-1)+6}{a-1} =2a+ \frac{3(a-1)}{a-1} + \frac{6}{a-1}= \\ \\ =2a+3+ \frac{6}{a-1}$
<span>

Q3. The answer is </span> $2 x^{2} +5x+2$ <span>

</span> $\frac{6 x^{3} +11 x^{2} -4x-4}{3x-2} = \frac{3x*2 x^{2}-2 x^{2} *2+15 x^{2} -4x-4 }{3x-2} = \frac{2 x^{2} (3x-2)+15 x^{2} -4x-4}{3x-2}= \\ \\ = \frac{2 x^{2} (3x-2)}{3x-2} + \frac{15 x^{2} -4x-4}{3x-2} =2 x^{2} +\frac{15 x^{2} -4x-4}{3x-2}=2 x^{2} + \frac{15 x^{2} -10x+6x-4}{3x-2}= \\ \\ =2 x^{2} + \frac{5x*3x-5x*2+6x-4}{3x-2} =2 x^{2} + \frac{5x(3x-2)+3x*2-2*2}{3x-2} = \\ \\ =2 x^{2} + \frac{5x(3x-2)}{3x-2} + \frac{3x*2-2*2}{3x-2} =2 x^{2} +5x+ \frac{2(3x-2)}{3x-2} =2 x^{2} +5x+2$
<span>

Q4. The answer is 2x + 7

</span> $\frac{6 x^{2} +11x-35}{3x-5} = \frac{6 x^{2} -10x+21x-35 }{3x-5} = \frac{3 x *2x-5*2x+7*3x-7*5 }{3x-5} = \\ \\ = \frac{2x(3x-5)+7(3x-5)}{3x-5}= = \frac{(3x-5)(2x+7)}{3x-5} =2x+7$
<span>

Q5. The answer is </span> $x+1- \frac{3}{x-1}$ <span>

</span> $\frac{ x^{2} -4}{x-1} = \frac{ x^{2} -x+x-1-3 }{x-1} = \frac{x*x-x+x-1-3}{x-1} = \frac{x(x-1)+(x-1)-3}{x-1} = \\ \\ \frac{(x+1)(x-1)-3}{x-1} = \frac{(x+1)(x-1)}{x-1} -\frac{3}{x-1} =x+1- \frac{3}{x-1}$

Q6. The answer is $y^{2} -2y+3$

$\frac{ y^{3}-4 y^{2}+7y-6 }{y-2} = \frac{y* y^{2} -2y^{2}-2 y^{2} +7y-6 }{y-2} = \frac{y^{2}(y-2)-2 y^{2} +7y-6}{y-2}= \\ \\ = \frac{y^{2}(y-2)}{y-2}+ \frac{-2 y^{2} +7y-6}{y-2} = y^{2} + \frac{-2 y^{2} +4y + 3y-6}{y-2} = \\ \\ =y^{2} + \frac{-2y*y-2y(-2)+3y-3*2}{y-2} = y^{2} + \frac{(-2y)(y-2)+3(y-2)}{y-2} = \\ \\ = y^{2} + \frac{(-2y+3)(y-2)}{y-2} = y^{2} +(-2y+3) =y^{2} -2y+3$
<span>

Q7. The answer is </span> $x^{2} +xy+ y^{2}}{x-y}$ <span>

</span> $\frac{ x^{3} - \frac{x}{y} y^{3} }{x-y} = \frac{(x-y)( x^{2} +xy+ y^{2}) }{x-y} = \frac{ x^{2} +xy+ y^{2}}{x-y}$
<span>

Q8. The answer is </span> $(a^{2} +2ab+2b^{2})$ <span>

</span> $\frac{a^{4} +4b^{4} }{a^{2}-2ab+2 b^{2} } = \frac{ (a^{2})^{2} +(2b)^{2}}{a^{2}-2ab+2 b^{2}} = \frac{(a^{2} -2ab+2b^{2})(a^{2} +2ab+2b^{2}) }{(a^{2} -2ab+2b^{2})} =(a^{2} +2ab+2b^{2})$ <span>

Q9. The answer is a^{n-8} - a^{-14}

</span> $\frac{ (a^{2}) ^{n} - a^{n-6} }{ a^{n+8} }= \frac{(a^{2}) ^{n} }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}} \\ \\ (x^{y}) ^{z}= x^{y*z} \\ \\ \frac{ x^{y} }{ x^{z} } = x^{y-z} \\ \\ 
\frac{(a^{2}) ^{n} }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}} =\frac{a^{2n} }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}} = a^{2n-(n+8)} - a^{n-6-(n+8)} = \\ \\ =a^{2n-n-8} - a^{n-6-n-8} = a^{n-8} - a^{-14}$

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3 years ago

The probability that aperson living in a certaincity owns a dog is estimated to be 0.3. Find the probabilitythat the tenth perso

allochka39001 [22]

The probability that the tenth person randomly interviewed in that city is the ﬁfth one to own a dog = 0.051

Given,

Consider the interview of a person is a trial.

Lets consider it success if a person owns a dog and consequently , a person without a dog will be a failure.'

Let the random variable X represent the number of persons to be interviewed to get 5 dog owner: in other words, to get 5 successors.

Probability of a success in each trial is p = 0.3. Therefore probability of a failure in each trial is q = 1 - p = 0.7 .

Because trials are independent. X has a negative binomial distribution with parameters k = 5, p = 0.3

P(X =x ) = b(x: k, p) = $[(x-1)(k-1)] p^kq^r^-^k$

where, x = k , k + 1 , k + 2......(1)

Now, Lets find the probability that the tenth person randomly interviewed in that city is the fifth one to own a dog

Using equation (1), we get:

P(X = 10) = b(10: 5, 0.3)

= [(10-1) (5 - 1)] $(0.3)^5(0.7)^1^0^-^5$

= [(9)(4)] $(0.3)^5(0.7)^1^0^-^5$

= 0.05146

Hence, The probability that the tenth person randomly interviewed in that city is the ﬁfth one to own a dog = 0.051

Learn more about Probability at:

brainly.com/question/743546

#SPJ9

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1 year ago

Lucas is offered either 15% or $21 off his total shopping bill. how much would have to be spent to make the 15% option the best

vladimir1956 [14]

Let the total bill be x. We have:
<span>0.15x > 21 <-- Since 15% is 0.15 using inequality calc. x </span>>$140

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3 years ago

If the factors of a quadratic function are (x + 2) and (x − 9), what are the x-intercepts of the function?

Zielflug [23.3K]

Answer: D. (-2,0) and (9,0)

Step-by-step explanation:

When you equate the factors to 0 you get

(X+2) = 0

X = -2

(X-9) = 0

X= 9

Here the Y value is 0 as you find the factors of x when the Y value is 0.

So the answer is (-2,0) and (9,0)

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3 years ago