# Math Help - A mixed bag of problems to solve

1. ## A mixed bag of problems to solve

I have a fair variety of problems that I am having some issues with--most I have worked out but need confirmation on, others I have no real idea what to do with.

The ones I need to double check are:

(1)
Assuming that none of the solutions are undefined,simplify each of the following:

(a) $\frac{a^2 b^2 c}{abc}$

My answer: Either abc or ab(I'm not too sure which, though I am leaning towards "ab")

(b) $\frac{a^{2m} b^{m+1}}{a^m b}$

My answer: $a^{2m-m}b^m$ or $a^mb^m$
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(2)
Simplify:
(a) $\frac{x-y}{x+y} + \frac{x+y}{x-y}$

My answer: $\frac{x-y}{x+y} + \frac{x+y}{x-y} \rightarrow \frac{(x-y)(x-y)}{(x+y)(x-y)} + \frac{(x+y)(x+y)}{(x+y)(x-y)}$ $\rightarrow \frac{(x-y)^2}{(x+y)(x-y)} + \frac{(x+y)^2}{(x+y)(x-y)} \rightarrow \frac{(x-y)^2+(x+y)^2}{(x+y)(x-y)} \rightarrow (x-y)+(x-y)$
__________________________________________________ ______________

(3)
If you are h feet above the earth's surface, the distance, d, that you can see is approximately d (in miles)=
$\sqrt{\frac{3h}{2}}$.
Suppose you are on the roof of a building 900 ft. tall. how far can you see?(calculate to two decimal places)

My answer: $h=900$ so $\sqrt{\frac{3h}{2}}=d \rightarrow \sqrt{\frac{3*900}{2}}=d \rightarrow \sqrt{\frac{2700}{2}}=d \rightarrow \sqrt{1250}=d \ \therefore \ d=36.74$ Thus, I can see for 36.74 Mi.
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(4)
According to Descartes' Rule of Signs, (a) how many positive real roots does each of the following have? (b) how many negative roots?

(1)
$f(a)=a^5-4a^2-7$
(a):1
(b):0

(2)
$f(x)=3x^3+9x^2+8x$
(a):0
(b):1
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Now for the ones I have no idea about.

(1)
Solve the equation
$x^3+9x^2+x+2=0$, if -2 is a root.
(I have tried to solve this one but when I do synthetic division on it I get the equation:
$x^2+x=0$ which I don't think will work)

(2)

Find the dimensions of a rectangle a
with the greatest area whose perimeter is 30 feet.
(I got nothing--I'm totally confused as to what to do with this one.)

Thank you!

2. Hi there,

In future please post one problem per thread, makes life easier for all concerned.

Originally Posted by MathBlaster47

The ones I need to double check are:

(1)
Assuming that none of the solutions are undefined,simplify each of the following:

(a) $\frac{a^2 b^2 c}{abc}$

My answer: Either abc or ab(I'm not too sure which, though I am leaning towards "ab")

$\frac{a\times a \times b\times b \times c}{a\times b \times c}$

After some cancelation you get $ab$
Originally Posted by MathBlaster47

(b) $\frac{a^{2m} b^{m+1}}{a^m b}$

My answer: $a^{2m-m}b^m$ or $a^mb^m$

You need to do it like this

$a^{2m-m}b^{m+1-1} =a^mb^m$

3. Originally Posted by pickslides
Hi there,

In future please post one problem per thread, makes life easier for all concerned.
Ok, I'll be sure to do that. I posted all of my questions in one place because I didn't want it to look like I was spamming the board.

4. Originally Posted by MathBlaster47
Ok, I'll be sure to do that. I posted all of my questions in one place because I didn't want it to look like I was spamming the board.
I understand your concern, thanks for that.

Maybe post 1 or 2 quesitons per thread, get the required help, solve them and come back later to post more.

5. How did you arrive at the last expression?

Originally Posted by MathBlaster47

__________________________________________________ ______________
(2)
Simplify:
(a) $\frac{x-y}{x+y} + \frac{x+y}{x-y}$

......

My answer: $\frac{(x-y)^2+(x+y)^2}{(x+y)(x-y)} \rightarrow (x-y)+(x-y)$
also if correct $(x-y)+(x-y) = 2 (x-y)$

6. Originally Posted by pickslides
How did you arrive at the last expression?

also if correct $(x-y)+(x-y) = 2 (x-y)$
I arrived at the last expression by expanding to $\frac{[(x-y)(x-y)]+[(x+y)(x+y)]}{(x-y)(x+y)}$ then I canceled (x-y) and (x+y) from the expression: [(x-y)(x-y)]+[(x+y)(x+y)]/(x-y)(x+y) became [(x-y)(x-y)]+[(x+y)(x+y)]/(x-y)(x+y) .

I guess I'm not quite getting it.
Where am I going wrong?

7. Originally Posted by MathBlaster47

(2)

Find the dimensions of a rectangle a
with the greatest area whose perimeter is 30 feet.
(I got nothing--I'm totally confused as to what to do with this one.)

Thank you!

This can be done in two ways.

The algebraic approach is as follows,
Let $x$ and $y$ be the sides of the rectangle in feet.

Therefore, the perimeter of the rectangle P = $2(x+y) = 30$

Now, the area of the rectangle A = $xy$

Now, since the Arithmetic mean of two numbers>= Their geometric mean,

$(x+y)/2$ >= $\sqrt{xy}$

Thus $0.25P >= 2\sqrt{A}$

Hope you can proceed from here.

8. Originally Posted by MathBlaster47
Solve the equation x^3+9x^2+x+2=0, if -2 is a root.
(I have tried to solve this one but when I do synthetic division on it I get the equation:
x^2+x=0 which I don't think will work)
x(x + 1) = 0
x = 0 or x + 1 = 0
OK?

9. Originally Posted by MathBlaster47
I arrived at the last expression by expanding to $\frac{[(x-y)(x-y)]+[(x+y)(x+y)]}{(x-y)(x+y)}$ then I canceled (x-y) and (x+y) from the expression.
1. Expand the brackets in the numerator:

$\frac{[(x-y)(x-y)]+[(x+y)(x+y)]}{(x-y)(x+y)} = \dfrac{x^2-2xy+y^2+x^2+2xy+y^2}{(x-y)(x+y)} =$ $\dfrac{2x^2+2y^2}{(x-y)(x+y)} =\dfrac{2(x^2+y^2)}{(x-y)(x+y)}$

2. You can't factor a sum of squares ( in $\mathbb{R}$)

3. In the dark ages of math you would have been boiled in oil for your "canceling"

10. Originally Posted by bandedkrait
This can be done in two ways.

The algebraic approach is as follows,
Let $x$ and $y$ be the sides of the rectangle in feet.

Therefore, the perimeter of the rectangle P = $2(x+y) = 30$

Now, the area of the rectangle A = $xy$

Now, since the Arithmetic mean of two numbers>= Their geometric mean,

$(x+y)/2$ >= $\sqrt{xy}$

Thus $0.25P >= 2\sqrt{A}$

Hope you can proceed from here.
Thank you!
Just so I'm clear: $15\geq \sqrt{A}$because $(x+y)/2 = (\frac{1}{4}) 30= 15$.
So: since $2x+2y=30$ and $xy=A$ then if $x=5$ and $y=10$, $2*5+2*10=30$ and $5*10=50$.
How do I make sure that the area I found is the greatest possible?

Originally Posted by Wilmer
x(x + 1) = 0
x = 0 or x + 1 = 0
OK?
Thank you muchly!

Originally Posted by earboth
1. Expand the brackets in the numerator:

$\frac{[(x-y)(x-y)]+[(x+y)(x+y)]}{(x-y)(x+y)} = \dfrac{x^2-2xy+y^2+x^2+2xy+y^2}{(x-y)(x+y)} =$ $\dfrac{2x^2+2y^2}{(x-y)(x+y)} =\dfrac{2(x^2+y^2)}{(x-y)(x+y)}$

2. You can't factor a sum of squares ( in $\mathbb{R}$)

3. In the dark ages of math you would have been boiled in oil for your "canceling"
1.Thank You!! I felt like something was off.....

2.Ah..... .

3.More oil? I haven't even gotten over the deep fryer I'm in right now......

11. Because the maximum value of the Geometric mean, $\sqrt{ab}$ is 15/2 = 7.5

This will happen for the situation $x=y$, i.e. when the rectangle is a square.

so using this, you get the maximum value of $x=15$

Hence find the maximum area, which is 225.