# questionable questions

• Feb 3rd 2010, 01:10 PM
NitroKnight
questionable questions
Greetings and salutations!

I have some questions that I can't figure out.

1
the perimeter of a rectangular yard is 290 feet. If its length is 25 feet greater than its width, what are the dimensions of the yard?

2
What is the equation in standard form of the line through (4,1) and (-2,3)?

3
Write in simplest form without negative exponents.
(2/3)^-2(4/5)^-1
[I think i have an answer to this one:
(2/3)^-2(4/5)^-1=(2^-2/3^-2)(5/4)=(2^-2*{1/3}^-2)(5/4)=({1/2}9)(5/4)=(9/2)(5/4)=48/8=6]

Thanks
• Feb 3rd 2010, 01:27 PM
Henryt999
Quote:

Originally Posted by NitroKnight
Greetings and salutations!

I have some questions that I can't figure out.

1
the perimeter of a rectangular yard is 290 feet. If its length is 25 feet greater than its width, what are the dimensions of the yard?
Set one side to be x and the other becomes x+25..
2
What is the equation in standard form of the line through (4,1) and (-2,3)?
You want to find the inclination of the line. That is given by: $\displaystyle \frac{\Delta y}{\Delta x}$.
$\displaystyle \Delta y = 1-3$
$\displaystyle \Delta x = 4-(-2) = 6$
now we have $\displaystyle y = \frac{-2}{6}\times$ $\displaystyle x = \frac{-1}{3}x$
Now you want to find the constant...ill leave that part to you
3
Write in simplest form without negative exponents.
(2/3)^-2(4/5)^-1
[I think i have an answer to this one:
(2/3)^-2(4/5)^-1=(2^-2/3^-2)(5/4)=(2^-2*{1/3}^-2)(5/4)=({1/2}9)(5/4)=(9/2)(5/4)=48/8=6]
Im not sure how this is menth to be cause I believe there are som bracetts missing.

Thanks

Hope it helps
• Feb 3rd 2010, 03:52 PM
NitroKnight
how do you make the math look like math and not gibberish like I did?
• Feb 3rd 2010, 10:28 PM
Henryt999
Hold the
Double click on the math text I have written and a window pops up showing how I have written.
• Feb 4th 2010, 10:04 AM
NitroKnight
$\displaystyle (2/3)^{-2}(4/5)^{-1}\rightarrow(2^{-2}/3^{-2})(5/4)\rightarrow(2^{-2}*{1/3}^{-2})(5/4)\rightarrow((1/2)9)(5/4)$$\displaystyle \rightarrow(9/2)(5/4)=48/8=6 This is about how it looked in my head and on my working paper, am I correct in my working? • Feb 4th 2010, 10:11 AM HallsofIvy Quote: Originally Posted by NitroKnight \displaystyle (2/3)^{-2}(4/5)^{-1}\rightarrow(2^{-2}/3^{-2})(5/4)\rightarrow(2^{-2}*{1/3}^{-2})(5/4)\rightarrow((1/2)9)(5/4)$$\displaystyle \rightarrow(9/2)(5/4)=48/8=6$

This is about how it looked in my head and on my working paper, am I correct in my working?

$\displaystyle (2/3)^2= 4/9$ so $\displaystyle (2/3)^{-2}= 9/4$. $\displaystyle (5/4)^{-1}= 4/5$ so $\displaystyle (5/2)^{-2}(5/4)^{-1}= (9/4)(4/5)= 9/5$.
• Feb 4th 2010, 10:12 AM
e^(i*pi)
Quote:

Originally Posted by NitroKnight
$\displaystyle (2/3)^{-2}(4/5)^{-1}\rightarrow(2^{-2}/3^{-2})(5/4)\rightarrow(2^{-2}*{1/3}^{-2})(5/4)\rightarrow((1/2)9)(5/4)$$\displaystyle \rightarrow(9/2)(5/4)=48/8=6$

This is about how it looked in my head and on my working paper, am I correct in my working?

That's still quite ambigious

$\displaystyle \left(\frac{2}{3}\right)^{-2} - 2\left(\frac{4}{5}\right)^{-1}$

With fractions raised to the negative power you can flip the fraction and change the sign of the exponent: $\displaystyle \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$

$\displaystyle \left(\frac{2}{3}\right)^{-2} - 2\left(\frac{4}{5}\right)^{-1} = \left(\frac{3}{2}\right)^2 - 2\left(\frac{5}{4}\right)$

When raising fractions to powers you need to apply the exponent to both the numerator and the denominator: $\displaystyle \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$

$\displaystyle \frac{9}{4} - \frac{10}{4} = -\frac{1}{4}$
• Feb 4th 2010, 03:21 PM
NitroKnight
Quote:

Originally Posted by HallsofIvy
$\displaystyle (2/3)^2= 4/9$ so $\displaystyle (2/3)^{-2}= 9/4$. $\displaystyle (5/4)^{-1}= 4/5$ so $\displaystyle (5/2)^{-2}(5/4)^{-1}= (9/4)(4/5)= 9/5$.

Quote:

Originally Posted by e^(i*pi)
That's still quite ambigious

$\displaystyle \left(\frac{2}{3}\right)^{-2} - 2\left(\frac{4}{5}\right)^{-1}$

With fractions raised to the negative power you can flip the fraction and change the sign of the exponent: $\displaystyle \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$

$\displaystyle \left(\frac{2}{3}\right)^{-2} - 2\left(\frac{4}{5}\right)^{-1} = \left(\frac{3}{2}\right)^2 - 2\left(\frac{5}{4}\right)$

When raising fractions to powers you need to apply the exponent to both the numerator and the denominator: $\displaystyle \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$

$\displaystyle \frac{9}{4} - \frac{10}{4} = -\frac{1}{4}$

Now I'm confused.....
Let me try to rewrite the original problem with my new found skills:

Write in the simplest form without negative exponents:
$\displaystyle (\frac{2}{3})^{-2}(\frac{4}{5})^{-1}$

My working
$\displaystyle (\frac{2}{3})^{-2}(\frac{4}{5})^{-1}\rightarrow(\frac{2^{-2}}{3^{-2}})(\frac{5}{4})\rightarrow(2^{-2})(\frac{1}{3^2})(\frac{5}{4})\rightarrow (\frac{1}{2}*\frac{1}{9})(\frac{5}{4}) \rightarrow (\frac{1}{18})(\frac{5}{4})\rightarrow\frac{5}{72}$ ?
I really think I went wrong here......What do I do?
• Feb 5th 2010, 05:56 AM
e^(i*pi)
Quote:

Originally Posted by NitroKnight
Now I'm confused.....
Let me try to rewrite the original problem with my new found skills:

Write in the simplest form without negative exponents:
$\displaystyle (\frac{2}{3})^{-2}(\frac{4}{5})^{-1}$

My working
$\displaystyle (\frac{2}{3})^{-2}(\frac{4}{5})^{-1}\rightarrow(\frac{2^{-2}}{3^{-2}})(\frac{5}{4})\rightarrow(2^{-2})(\frac{1}{3^2})(\frac{5}{4})\rightarrow (\frac{1}{2}*\frac{1}{9})(\frac{5}{4}) \rightarrow (\frac{1}{18})(\frac{5}{4})\rightarrow\frac{5}{72}$ ?
I really think I went wrong here......What do I do?

You'd said that $\displaystyle \frac{1}{3^{-2}} = \frac{1}{9}$. That expression is equal to 9: $\displaystyle \frac{1}{\frac{1}{9}} = 9$

You can flip the expression inside the brackets for any power as long as you change the sign:

$\displaystyle \left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^2 = \frac{3^2}{2^2} = \frac{9}{4}$
• Feb 5th 2010, 08:35 AM
NitroKnight
$\displaystyle (\frac{2}{3})^{-2}(\frac{4}{5})^{-1}\rightarrow(\frac{3}{2})^2(\frac{5}{4})\rightarr ow(\frac{9}{4})(\frac{5}{4})\rightarrow \frac{9*5}{4*4}\rightarrow\frac{45}{16}$ ?