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galben [10]
3 years ago
12

Combining like terms of 10x - 3+ 11 - 2x + 3x, simplifies the expression to​

Mathematics
2 answers:
lesya692 [45]3 years ago
8 0

Answer:

-9x+8

Step-by-step explanation:

balu736 [363]3 years ago
7 0

15x - 3+11 should be the answer

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5. Which line appears to have an x-intercept of -5 and a y-intercept of 3?
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Use the power series method to solve the given initial-value problem. (Format your final answer as an elementary function.)
olya-2409 [2.1K]

You're looking for a solution of the form

\displaystyle y = \sum_{n=0}^\infty a_n x^n

Differentiating twice yields

\displaystyle y' = \sum_{n=0}^\infty n a_n x^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n

\displaystyle y'' = \sum_{n=0}^\infty n(n-1) a_n x^{n-2} = \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n

Substitute these series into the DE:

\displaystyle (x-1) \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n - x \sum_{n=0}^\infty (n+1) a_{n+1} x^n + \sum_{n=0}^\infty a_n x^n = 0

\displaystyle \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^{n+1} - \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n \\\\ \ldots \ldots \ldots - \sum_{n=0}^\infty (n+1) a_{n+1} x^{n+1} + \sum_{n=0}^\infty a_n x^n = 0

\displaystyle \sum_{n=1}^\infty n(n+1) a_{n+1} x^n - \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n \\\\ \ldots \ldots \ldots - \sum_{n=1}^\infty n a_n x^n + \sum_{n=0}^\infty a_n x^n = 0

Two of these series start with a linear term, while the other two start with a constant. Remove the constant terms of the latter two series, then condense the remaining series into one:

\displaystyle a_0-2a_2 + \sum_{n=1}^\infty \bigg(n(n+1)a_{n+1}-(n+1)(n+2)a_{n+2}-na_n+a_n\bigg) x^n = 0

which indicates that the coefficients in the series solution are governed by the recurrence,

\begin{cases}y(0)=a_0 = -7\\y'(0)=a_1 = 3\\(n+1)(n+2)a_{n+2}-n(n+1)a_{n+1}+(n-1)a_n=0&\text{for }n\ge0\end{cases}

Use the recurrence to get the first few coefficients:

\{a_n\}_{n\ge0} = \left\{-7,3,-\dfrac72,-\dfrac76,-\dfrac7{24},-\dfrac7{120},\ldots\right\}

You might recognize that each coefficient in the <em>n</em>-th position of the list (starting at <em>n</em> = 0) involving a factor of -7 has a denominator resembling a factorial. Indeed,

-7 = -7/0!

-7/2 = -7/2!

-7/6 = -7/3!

and so on, with only the coefficient in the <em>n</em> = 1 position being the odd one out. So we have

\displaystyle y = \sum_{n=0}^\infty a_n x^n \\\\ y = -\frac7{0!} + 3x - \frac7{2!}x^2 - \frac7{3!}x^3 - \frac7{4!}x^4 + \cdots

which looks a lot like the power series expansion for -7<em>eˣ</em>.

Fortunately, we can rewrite the linear term as

3<em>x</em> = 10<em>x</em> - 7<em>x</em> = 10<em>x</em> - 7/1! <em>x</em>

and in doing so, we can condense this solution to

\displaystyle y = 10x -\frac7{0!} - \frac7{1!}x - \frac7{2!}x^2 - \frac7{3!}x^3 - \frac7{4!}x^4 + \cdots \\\\ \boxed{y = 10x - 7e^x}

Just to confirm this solution is valid: we have

<em>y</em> = 10<em>x</em> - 7<em>eˣ</em>   ==>   <em>y</em> (0) = 0 - 7 = -7

<em>y'</em> = 10 - 7<em>eˣ</em>   ==>   <em>y'</em> (0) = 10 - 7 = 3

<em>y''</em> = -7<em>eˣ</em>

and substituting into the DE gives

-7<em>eˣ</em> (<em>x</em> - 1) - <em>x</em> (10 - 7<em>eˣ </em>) + (10<em>x</em> - 7<em>eˣ</em> ) = 0

as required.

8 0
3 years ago
Score on last try: 0 of 8 pts. See Details for more.
Agata [3.3K]

Answer:

there are 4 aces in the deck of 52 cards

so the probability to draw one ace is 4/52

if we draw an ace, there are 3 left in 51 cards.

so the next probability will be 3/51

we need to multiply probabilities to know how likely it is that this happens in a row.

for three in a row that's

\frac{4}{52}  \times  \frac{3}{51} \times  \frac{2}{50}

rewritten its:

\frac{4 \times 3 \times 2}{52 \times 51 \times 50}

we'll see soon why this is easier to write.

we also have to account for the two cards that are no aces. it doesn't matter if the first two cards are no aces, the last two, or any rodney. just the ammount of non-aces is important.

there are 48 non-aces in the deck.

so the probability to draw one of them in the first step is 48/52.

wich each draw the number of cards left in the deck decreases by 1. and depending what we draw, that ammount also decreases.

we draw 5 cards, so we know what the denominators will be. the numerator will be the rest-ammounts of each event, like shown in the complicated fraction above.

let's write it most easily as this:

\frac{4 \times 3 \times 2 \times 48 \times 47}{52 \times 51 \times 50 \times 49 \times 48}

this scenario looks like it only describes 3 aces, then two non-aces, but the order of multiplication isn't important. you could put the 48 and 47 to the left, same solution. so this expression represents all permutations of "3 aces when drawing 5 cards", including the two cards that are not aces.

that's why writing it as one convoluted fraction is indeed better and simpler than many fraction multiplied in row.

time to grap a calculator.

our probability for the event is

54144 possible valid outcomes out of

311875200 possible combinations.

luckily the calculator can reduce that to a simpler form of

\frac{47}{270725}

(reduced with 2^6*3^, big number, but tried it out by hand stacking up prime factors until hittinga prime in the numerator)

written as a readable kind of number you probably know and like it's

0.0001736079047

in decimal form. now rounded as asked for:

0.00017

(as percentage: 0.017%)

hope the way how we got here is somewhat clear. feel free to ask any questions.

kind regards

Alex

6 0
3 years ago
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