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2 years ago

Orders of operations. 2{6+2(6*7-18)}=

Mathematics

1 answer:

Over [174]2 years ago

8 0

6*7 = 42
42-18= 24

6+2
8*24=192

2(192)= 384

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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

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

4 0

1 year ago

If f(x)=1/x+2 what is f(3+h)?

Wittaler [7]

I don't know if you want the whole x+2 on the bottom or not, but I will put it on the bottom for now.
Replace x with (3+h)

f(3+h)=1/[(3+h)+2]

Now if it's supposed to be JUST x in the denominator...
Replace x with (3+h)

f(3+h)=1/(3+h)+2

Best wishes!

3 0

2 years ago

here is the line for the question J’Wynn rides his bike 2 miles each day. What equation can be used to find r,

Irina18 [472]

The answer to this question is C.

7 0

2 years ago

HELP ASAP IM IN A TIMED TEST 8TH GRADE MATH

sattari [20]

Answer:

Step-by-step explanation:

Convert them all to the form y = mx + c

4x + 3y = 15

3y = -4x + 15

y = -4/3 x + 5

3x - 4y = -8

-4y = -3x - 8

4y = 3x + 8

y = 3/4x + 2

y + 1 = 4/3(x - 6)

y + 1 = 4/3x - 8

y = 4/3x - 9

y = 3/4x - 5

8 0

2 years ago

If f(x) is an antiderivative of g(x), and g(x) is an antiderivative of h(x), then

Goryan [66]

If f(x) is an anti-derivative of g(x), then g(x) is the derivative of f(x). Similarly, if g(x) is the anti-derivative of h(x), then h(x) must be the derivative of g(x). Therefore, h(x) must be the second derivative of f(x); this is the same as choice A.

I hope this helps.

7 0

2 years ago