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UkoKoshka [18]
3 years ago
10

20 points ,. will mark as brainlyest Which is a solution to x + 5 = 8?

Mathematics
2 answers:
tiny-mole [99]3 years ago
6 0

Answer:

3

Step-by-step explanation:

8-5=3

nadya68 [22]3 years ago
5 0

Answer:

3

Step-by-step explanation:

3 plus 5 equals 8

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If 5x - 2 = -12, then x =
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Answer:

x=-2

Step-by-step explanation:

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3 years ago
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Katherine has $140 in the bank and is saving $6 per week. Abbie has $462 in the bank, but spending at a rate of $10m per week. W
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140 + 6x = 462 - 10x

Hope I helped

Have a great day

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2 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,

5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0

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}}

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2 years ago
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QveST [7]
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