The libary has at least 5,000 books witch inequality represent the number of books,b,at the library

Solve and show steps. will award brainliest.

a_sh-v [17]

For the first question please find the attached diagram.

As per the diagram, P is the upstream point and Q is the downstream point. The distance between P and Q is 22.5 miles.

Let the speed of the boat in the still waters of the lake be represented by S.

Then, when the boat travels upstream, the net speed of the boat will be (S-6) miles per hour because the river flows downstream and thus the speed of the boat will have to be subtracted from the speed of the river.

Now, we know that the relationship between the net speed, distance and time of travel is give as:

Distance = Net Speed x Time of travel

For the upstream ride of the board we know that Distance is 22.5 miles and Net Speed is (S-6). Therefore, the above equation will become:

$22.5=(S-6)\times T_{1}$ where $T_{1}$ represents the time taken to travel upstream.

We can rearrange the above equation to be:

$T_{1}=\frac{22.5 }{S-6}$ ......................(Equation 1)

By similar arguments we know that the downstream speed of the boat is S+6 and the distance travelled is the same and so the time taken to travel downstream (represented by $T_{2}$ ) will be:

$T_{2}=\frac{22.5}{S+6}$ ................(Equation 2)

Now, we know that the total time of travel should be 9 hours.

This means that: $T_{1}+T_{2}=9$ ............(Equation 3)

Plugging in the values of $T_{1}$ and $T_{2}$ from (Equation 1) and (Equation 2) into (Equation 3), we get:

$\frac{22.5 }{S-6} +\frac{22.5 }{S+6}=9$

Simplifying the above we will get a quadratic equation:

$9S^2-45S-54=0$

The roots of this quadratic equation are:

$S=-1$ and $S=6$

Since, speed cannot be negative, $S=-1$ is out of consideration.

The speed of the boat in the lake is thus $S=6$ miles per hour.

But we have a problem with S=6 too. The problem is that if S=6, then the boat will not be able to move upstream.

Let us solve problem 2

We are given that: $\frac{x-2}{x+3}+\frac{10x}{x^2-9}$

We can rewrite it as:

$\frac{x-2}{x+3}+\frac{10x}{(x-3)(x+3)}$

$\frac{(x-3)(x-2)+10x}{(x-3)(x+3)} =\frac{x^2-5x+6+10x}{(x-3)(x+3)}$

Now, the numerator can be simplified as:

$\frac{x^2+10x+6}{(x-3)(x+3)} =\frac{(x+3)(x+2)}{(x-3)(x+3)} =\frac{x+2}{x-3}$

Thus, our final simplified answer is:

$\frac{x+2}{x-3}$

The restriction on the variable x is that it cannot be equal to either +3 or -3 as that would make the denominator of the original question equal to zero.

Thus, the restriction is $x\neq \pm 3$