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BabaBlast [244]
4 years ago
14

How would you solve this: x^2-9x+18

Mathematics
2 answers:
Ostrovityanka [42]4 years ago
7 0
Hope this helps you!!!

gulaghasi [49]4 years ago
7 0

Hello!

To factor this, we will want to find the GCF of all of these numbers. It cannot involve x, as 18 is a constant.

x·x-9x+18

As you can see, there is no common factor, so we cannot multiply a parentheses by one number. What we can do, though, is multiply two parentheses. We can use the FOIL method, which means that to unfactor two parentheses, you multiply the first numbers, outside, inside, and last. Let's try doing so.

(a,b)(y,z)

First of all, a multiplied by x will probably have to equal x², as it appears first. This means a and x are x.

(x,b)(x,z)

As you can see, the outside and inside are both going to have an x in them. This means that we will want some numbers that will add to make -9, and will multiply to make 18. If you think about it, -3 and -6 would work.

-3+-6=-9

-3(-6)=18

Therefore, our factored answer is below.

(x-3)(x-6)

I hope this helps!


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4,005 is divisible by 2? True or false?
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Leah drives 45 miles per hour for 2 hours then 32 miles per hour for 90 minutes . What is her average speed
Citrus2011 [14]

Answer:

 =39.42857143 miles per hour

Step-by-step explanation:

To find the average speed, we need to the total distance and the total time

Distance 1

45 miles for 2 hours

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90 minutes * 1 hour/60 minutes = 1.5 hours

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3 0
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Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand
dybincka [34]

Answer: Part a) P(a)=\frac{1}{\binom{r+w}{r}}

part b)P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = P(E)=\frac{Favourablecases}{TotalCases}

For part a)

Probability that a red ball is drawn in first attempt = P(E_{1})=\frac{r}{r+w}

Probability that a red ball is drawn in second attempt=P(E_{2})=\frac{r-1}{r+w-1}

Probability that a red ball is drawn in third attempt = P(E_{3})=\frac{r-2}{r+w-1}

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = P(E_{i})=\frac{r-i}{r+w-i}

Thus the probability that events E_{1},E_{2}....E_{i} occur in succession is

P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...

Thus P(E)=\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!

Thus our probability becomes

P(E)=\frac{1}{\binom{r+w}{r}}

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at (r+1)^{th} draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals \binom{r}{r-1}

Thus the probability becomes P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}

Thus required probability of case b becomes P(E)+ P(E_{i})

= P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\

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