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3 years ago

Please help anyone if you can help thanks

Mathematics

1 answer:

-BARSIC- [3]3 years ago

7 0

Answer:

t=5/10n

Step-by-step explanation:

The question wants to know how much time it takes to run one lap, not how many laps can be run in a certain amount of time.

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Solve the given initial-value problem. the de is of the form dy dx = f(ax + by + c), which is given in (5) of section 2.5. dy dx

shutvik [7]

$\dfrac{\mathrm dy}{\mathrm dx}=\cos(x+y)$

Let $v=x+y$ , so that $\dfrac{\mathrm dv}{\mathrm dx}-1=\dfrac{\mathrm dy}{\mathrm dx}$ :

$\dfrac{\mathrm dv}{\mathrm dx}=\cos v+1$

Now the ODE is separable, and we have

$\dfrac{\mathrm dv}{1+\cos v}=\mathrm dx$

Integrating both sides gives

$\displaystyle\int\frac{\mathrm dv}{1+\cos v}=\int\mathrm dx$

For the integral on the left, rewrite the integrand as

$\dfrac1{1+\cos v}\cdot\dfrac{1-\cos v}{1-\cos v}=\dfrac{1-\cos v}{1-\cos^2v}=\csc^2v-\csc v\cot v$

Then

$\displaystyle\int\frac{\mathrm dv}{1+\cos v}=-\cot v+\csc v+C$

and so

$\csc v-\cot v=x+C$

$\csc(x+y)-\cot(x+y)=x+C$

Given that $y(0)=\dfrac\pi2$ , we find

$\csc\left(0+\dfrac\pi2\right)-\cot\left(0+\dfrac\pi2\right)=0+C\implies C=1$

so that the particular solution to this IVP is

$\csc(x+y)-\cot(x+y)=x+1$

5 0

3 years ago

If 8 boys went to the washroom, there would be as many boys as girls in the room. 3 If 8 girls went to the washroom, there would

Delicious77 [7]

Answer:

24 kids in the classroom.

5 0

3 years ago

Sean has a piece of string 7/8 meter long. He uses 1/5 of the piece of string to tie a package and cuts the rest into 5 equal pi

Olin [163]

Let change 7/8 and 1/5 into decimal point so 0.88 and 0.2
0.88-0.2 =0.68
then we want 5 equal length so we divided
0.68 / 5 =0.14

6 0

2 years ago

How do you do series with an n

SOVA2 [1]

The problem can be solved step by step, if we know certain basic rules of summation. Following rules assume summation limits are identical.
$\sum{a+b}=\sum{a}+\sum{b}$
$\sum{kx}=k\sum{x}$
$\sum_{r=1}^n{1}=n$
$\sum_{r=1}^n{r}=n(n+1)/2$

Armed with the above rules, we can split up the summation into simple terms:
$\sum_{r=1}^n{40r-21n+8}=n$
$=40\sum_{r=1}^n{r}-21n\sum_{r=1}^n{1}+8\sum_{r=1}^n{1}$
$=40\frac{n(n+1)}{2}-21n^2+8n$
$=20n(n+1)-21n^2+8n$
$=28n-n^2$

=> (a) f(x)=28n-n^2
=> f'(x)=28-2n
=> at f'(x)=0 => x=14
Since f''(x)=-2 <0 therefore f(14) is a maximum
(b) f(x) is a maximum when n=14

(c) the maximum value of f(x) is f(14)=196

4 0

2 years ago

Given: BC bisects DBE. Prove: ABD is congruent to ABE

Arada [10]

Answer:

See below

Step-by-step explanation:

BC bisects <DBE and if AC is a straight line then you have that:

<ABD + <DBC = 180 (straight angle because of line AC)

<ABE + <CBE = 180 ( straight angle because of line AC)

Because BC bisects <DBE => < DBC = <CBE

So <ABD and <ABE must be the same to both sum 180 when added < DBC

7 0

3 years ago