Math Help - questionable questions

1. 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

2. 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: $\frac{\Delta y}{\Delta x}$.
$\Delta y = 1-3$
$\Delta x = 4-(-2) = 6$
now we have $y = \frac{-2}{6}\times$ $x = \frac{-1}{3}x$
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

3. how do you make the math look like math and not gibberish like I did?

4. Hold the

Double click on the math text I have written and a window pops up showing how I have written.

5. $(2/3)^{-2}(4/5)^{-1}\rightarrow(2^{-2}/3^{-2})(5/4)\rightarrow(2^{-2}*{1/3}^{-2})(5/4)\rightarrow((1/2)9)(5/4)$ $\rightarrow(9/2)(5/4)=48/8=6$

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

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

This is about how it looked in my head and on my working paper, am I correct in my working?
$(2/3)^2= 4/9$ so $(2/3)^{-2}= 9/4$. $(5/4)^{-1}= 4/5$ so $(5/2)^{-2}(5/4)^{-1}= (9/4)(4/5)= 9/5$.

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

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

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

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

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

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

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

8. Originally Posted by HallsofIvy
$(2/3)^2= 4/9$ so $(2/3)^{-2}= 9/4$. $(5/4)^{-1}= 4/5$ so $(5/2)^{-2}(5/4)^{-1}= (9/4)(4/5)= 9/5$.
Originally Posted by e^(i*pi)
That's still quite ambigious

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

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

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

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

$\frac{9}{4} - \frac{10}{4} = -\frac{1}{4}$
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:
$(\frac{2}{3})^{-2}(\frac{4}{5})^{-1}$

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

9. 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:
$(\frac{2}{3})^{-2}(\frac{4}{5})^{-1}$

My working
$(\frac{2}{3})^{-2}(\frac{4}{5})^{-1}\rightarrow(\frac{2^{-2}}{3^{-2}})(\frac{5}{4})\rightarrow(2^{-2})(\frac{1}{3^2})(\frac{5}{4})\rightarrow (\frac{1}{2}*\frac{1}{9})(\frac{5}{4}) \rightarrow (\frac{1}{18})(\frac{5}{4})\rightarrow\frac{5}{72}$ ?
I really think I went wrong here......What do I do?
You'd said that $\frac{1}{3^{-2}} = \frac{1}{9}$. That expression is equal to 9: $\frac{1}{\frac{1}{9}} = 9$

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

$\left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^2 = \frac{3^2}{2^2} = \frac{9}{4}$

$(\frac{2}{3})^{-2}(\frac{4}{5})^{-1}\rightarrow(\frac{3}{2})^2(\frac{5}{4})\rightarr ow(\frac{9}{4})(\frac{5}{4})\rightarrow \frac{9*5}{4*4}\rightarrow\frac{45}{16}$ ?