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Illusion [34]
3 years ago
13

Simplify 2x - 8x - 6y + 9 -14 + x + 11y. Thanks!

Mathematics
2 answers:
yarga [219]3 years ago
7 0
-5x+5y-5 is the answer
denis23 [38]3 years ago
6 0
2x-8x-6y+9-14+x+11y    2x-8x=-6x
-6x-6y+9-14+x+11y        9-14=-5
-6x-6y-5+x+11y             -6x+x=-5x
-5x-6y-5+11y                 -6y+11y=5y
-5x+5y-5    
hope this helps 
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a piece of paper is 0.0075 in thick how many sheets of paper will be in a stack that is 2.25 inches high
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3 years ago
(1) Which of the following is the factored form of:<br> x2 + 10x +9
Degger [83]

Answer:

( + 1 ) ( + 9 )

Step-by-step explanation:

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3 years ago
Please help me with the below question.
VMariaS [17]

By letting

y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}

we get derivatives

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

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

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0

Examine the lowest degree term \left(x^{r-1}\right), which gives rise to the indicial equation,

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with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients c_k is

(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}

so that with r = 4/5, the coefficients are governed by

c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}

c) Starting with c_0=1, we find

c_1 = -\dfrac{c_0}5 = -\dfrac15

c_2 = -\dfrac{c_1}{10} = \dfrac1{50}

so that the first three terms of the solution are

\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}

4 0
2 years ago
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