All categories

2 years ago

What is the total number of different 9-letter arrangements that can

Mathematics

1 answer:

MissTica2 years ago

5 0

F is it67 17 and 9 and 78

You might be interested in

Please help me please ! Quick

UkoKoshka [18]

Answer:

4,2 is tge answer due to the interaction point plus it says the solution is at the interaction.

8 0

3 years ago

Read 2 more answers

The local newspaper has letters to the editor from 10 people. If this number represents 2% of all of the newspaper's readers,

RideAnS [48]

Answer:

The newspaper has 500 readers.

Step-by-step explanation:

We have that:

10 people is 2$ of all newspapers readers, which total t. This means that:

$0.02t = 10$

$t = \frac{10}{0.02}$

$t = \frac{1000}{2}$

$t = 500$

The newspaper has 500 readers.

5 0

3 years ago

Consider the implicit differential equation <img src="https://tex.z-dn.net/?f=%2849%20y%5E%7B3%7D%20%2B%2045%20xy%29%20dx%20%2B%

BaLLatris [955]

We're looking for an integrating factor $\mu(x,y)=x^py^q$ such that

$\mu\underbrace{(49y^3+45xy)}_M\,\mathrm dx+\mu\underbrace{(98xy^2+50x^2)}_N\,\mathrm dy=0$

is exact, which would require that

$(\mu M)_y=(\mu N)_x$
$(49x^py^{q+3}+45x^{p+1}y^{q+1})_y=(98x^{p+1}y^{q+2}+50x^{p+2}y^q)_x$
$49(q+3)x^py^{q+2}+45(q+1)x^{p+1}y^q=98(p+1)x^py^{q+2}+50(p+2)x^{p+1}y^q$
$\implies\begin{cases}49(q+3)=98(p+1)\\45(q+1)=50(p+2)\end{cases}\implies p=\dfrac52,q=4$

You can verify that $(\mu M)_y=(\mu N)_x$ if you'd like. With the ODE now exact, we have a solution $F(x,y)=C$ such that

$F_x=\mu M$
$F=\displaystyle\int(49y^3+45xy)x^{5/2}y^4\,\mathrm dx$
$F=10x^{9/2}y^5+14x^{7/2}y^7+f(y)$

$F_y=\mu N$
$50x^{9/2}y^4+98x^{7/2}y^6+f'(y)=98x^{7/2}y^2+50x^{9/2}y^4$
$f'(y)=0$
$\implies f(y)=C$

and so the general solution is

$F(x,y)=10x^{9/2}y^5+14x^{7/2}y^7=C$

8 0

3 years ago

It costs $12 to place 8 lines of copy in a local advertising circular . What is the unit rate per line

Sauron [17]

About $1.5 per line

7 0

3 years ago

How many different positive integers can be represented as a difference of two distinct members of the set {1, 2, 3, 4, 5, 6}?

slega [8]

For the difference to be positive, when both the minuend and the subtrahend are positive, the former must be greater than the latter.

Therefore

minuend = 2 ⇒ subtrahend = 1 ← 1 number

minuend = 3 ⇒ subtrahend = 1,2 ← 2 numbers

minuend = 4 ⇒ subtrahend = 1,2,3 ← 3 numbers

minuend = 5 ⇒ subtrahend = 1,2,3,4 ← 4 numbers

minuend = 6 ⇒ subtrahend = 1,2,3,4,5 ← 5 numbers

1+2+3+4+5=<u>15</u>

5 0

3 years ago