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Alenkinab [10]
1 year ago
12

H = ac + mn solve for n

Mathematics
1 answer:
notka56 [123]1 year ago
6 0

Answer:

n =  \frac{h - ac}{m}

Step-by-step explanation:

h = ac + mn

mn = h - ac

n =  \frac{h - ac}{m}

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x = c1 cos(t) + c2 sin(t) is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the seco
igomit [66]

Answer:

x=-cos(t)+2sin(t)

Step-by-step explanation:

The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:

A differential equation of the form:

a_n y^n +a_n_-_1y^{n-1}+...+a_1y'+a_oy=0

Will have a characteristic equation of the form:

a_n r^n +a_n_-_1r^{n-1}+...+a_1r+a_o=0

Where solutions r_1,r_2...,r_n are the roots from which the general solution can be found.

For real roots the solution is given by:

y(t)=c_1e^{r_1t} +c_2e^{r_2t}

For real repeated roots the solution is given by:

y(t)=c_1e^{rt} +c_2te^{rt}

For complex roots the solution is given by:

y(t)=c_1e^{\lambda t} cos(\mu t)+c_2e^{\lambda t} sin(\mu t)

Where:

r_1_,_2=\lambda \pm \mu i

Let's find the solution for x''+x=0 using the previous information:

The characteristic equation is:

r^{2} +1=0

So, the roots are given by:

r_1_,_2=0\pm \sqrt{-1} =\pm i

Therefore, the solution is:

x(t)=c_1cos(t)+c_2sin(t)

As you can see, is the same solution provided by the problem.

Moving on, let's find the derivative of x(t) in order to find the constants c_1 and c_2:

x'(t)=-c_1sin(t)+c_2cos(t)

Evaluating the initial conditions:

x(0)=-1\\\\-1=c_1cos(0)+c_2sin(0)\\\\-1=c_1

And

x'(0)=2\\\\2=-c_1sin(0)+c_2cos(0)\\\\2=c_2

Now we have found the value of the constants, the solution of the second-order IVP is:

x=-cos(t)+2sin(t)

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3 years ago
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What is 8 over 9 in ratio
fenix001 [56]
8 over 9
= 8/9
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4 0
3 years ago
Read 2 more answers
Twenty different books are to be put on five book shelves, each of which holds at least twenty books.
olya-2409 [2.1K]

Answer:

(a) 10,626 different arrangements

(b) 95,367,431,640,625 different arrangements

(c) 2.5852017 × 10²² different arrangements

Step-by-step explanation:

(a) How many different arrangements are there if you only care about the number of books on the shelves (and not which book is where)?

We use the combination formula for this

C(n , r) = n + r - 1C r - 1

n = 20

r = 5

= 20 + 5 - 1 C 5-1

= 24C4

= 24!/4 ! × (24 - 4)

= 24!/4! × 20!

= 10,626 arrangements

(b) How many different arrangements are there if you care about which books are where, but the order of the books on the shelves doesn't matter?

Since the order of the books on the shelves does not matter,

The calculation is given as

5²⁰ = 95,367,431,640,625 arrangements

(c) How many different arrangements are there if the order on the shelves does matter?

Since order matters now

Step 1

We use the combination formula for this

C(n , r) = n + r - 1C r - 1

n = 20

r = 5

= 20 + 5 - 1 C 5-1

= 24C4

= 24!/4 ! × (24 - 4)

= 24!/4! × 20!

= 10,626

Step 2

We find the factorial of the number books

= 20!

= 2,432,902,008,176,640,000

Step 3

The different arrangements there are if the order on the shelves does matter is calculated

= 10,626 × 2,432,902,008,176,640,000

= 2.5852017 × 10²² different arrangements

8 0
3 years ago
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