All categories

2 years ago

I NEED HELP PLEASE, THANKS! :)

Mathematics

2 answers:

Aleonysh [2.5K]2 years ago

5 0

but where is ur question ??????

Gala2k [10]2 years ago

4 0

Answer:

where is the question?

Step-by-step explanation:

You might be interested in

A gymnasium can hold no more than 650 people. A permanent permanent bleacher in the gymnasium holds 136 people. The event organi

sattari [20]

136 + 25x ≤ 650
136 - 136 + 25x ≤ 650 - 136 
25x ≤ 514 
(1/25)(25x) ≤ (514)(1/25) 
x ≤ 20.56 
there can be at at the most, 20 chair in each row.

8 0

3 years ago

Read 2 more answers

Envision geometry 2-4 lesson quiz answer key

hoa [83]

Answer:

-2 i think im not to suure

Step-by-step explanation:

6 0

1 year ago

Malco weighs 42 more pounds than Peter the sum of their Weights is 286 pounds how much does Peter way

Travka [436]

122
286-42=144
144\2=122
122(peter)+164(malco)=286(total weight)

7 0

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

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

4 0

1 year ago

Please help me this is very important its a test due in 30 minutes

Mashcka [7]

A
Please mark as a brainliest

5 0

2 years ago