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1 year ago

3b-2a Need help on assignment

Mathematics

1 answer:

Bond [772]1 year ago

8 0

Answer: -2a + 3b

Solution:

3b - 2a
Simplify

= -2a + 3b

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What the best answer please help

Karo-lina-s [1.5K]

The answer B is correct, as this is the floor function. The floor function takes the greatest integer below the number, so that's why B is correct.

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2 years ago

Please help!I will make brainliest

forsale [732]

It would be the first one

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2 years ago

Read 2 more answers

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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1 year ago

The difference of a number and its successor

Arisa [49]

Answer:

Step-by-step explanation:

The difference between a number and its successor is 1

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2 years ago

I NEED HELP ON THIS SO BAD WILL GIVE CROWN

Stells [14]

5.5 quarts equals 22 cups. (There are 4cups in a quart, so 4 * 5.5 = 22)

22-4= 18 cups

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2 years ago

3b-2a Need help on assignment ​

3b-2a Need help on assignment