SOLVE. x cubed over x squared - 4 divided by x cubed - 9x squared over x squared + 7x -18

2 answers:

7 0

x^3/x^2-4/x^3-9x^2/x^2+7x-18
factor everything so you know what to cancel out; Also, take the bottom "fraction"; turn it over, spank it, and multiply it to the top "fraction"; like this:
xxx (x+9)(x-2)
--------- * ----------
(x-2)(x+2) xx (x-9)

Now cancel like terms top to bottom
x(x+9)
----------
(x+2) (x-9)

Tom [10]3 years ago

3 0

$\frac{ x^{3} }{ x^{2} -4}: \frac{ x^{3} -9 x^{2} }{ x^{2} +7x-18}= \\ = \frac{ x^{3} }{(x+2)(x-2)}* \frac{ x^{2} +7x-18}{ x^{2} (x-9)}= \\ = \frac{ x^{3} }{(x+2)(x-2)}* \frac{(x-2)(x+9)}{ x^{2} (x-9)}= \\ = \frac{x(x+9)}{(x+2)(x-9)}$

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Answer:

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Step-by-step explanation:

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The real part is the number that stands alone, while the imaginary part is the number that is directly beside the i.

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stiks02 [169]

Answer:

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6 0

3 years ago

you are rolling a 10-sided die with sides numbered 1-10. you keep rolling the die until you roll a prime number and then stop. l

evablogger [386]

The formula for E(x) would be SUM[ X(n)P(n)] from n = 1 to n = infinity until we get the prime number

In the given problem, a 10-sided die with 1-10 numbers is rolled until we get a prime number

Prime numbers between 1-10 = 2,3,5,7 (4 prime numbers)

Non - prime numbers = 1,4,6,8,9,10( 6 non-prime number)

Let X be the number of times the die is rolled

The probability of getting a prime =

$P(Prime) = \frac{4}{10}$

Now, the value of

E(X)=∑x.P(x) [ x-> {1, infinity}]

= P(1)X(1) +P(2)X(2)+P(3)X(3)+P(4)X(4)........+ P(n)X(n)

= 1. $\frac{4}{10}$ + 2. $\frac{6}{10}$ . $\frac{4}{10}$ + 3. $\frac{6}{10}$ . $\frac{6}{10}$ $\frac{4}{10}$ + 4. $\frac{6}{10}$ . $\frac{6}{10}$ . $\frac{6}{10}$ . $\frac{4}{10}$ .......