math...HELP PLEASE!!!?

Question

a seven foot ladder, leaning agains a wall, touches the wall x feet above the ground. write an expression (in terms of x) for the distance from the foot of the ladder to the base of the wall.

i got f(x)=sqrt(x^2 +49)

is that right??



and i need clarification on a previously asked question....
simplify using only positive exponents. do not rationalize the denominator.

1. square root (4x-16) / fourth root {(x-4)^3}

2. [ (1 / x^-2) + (4 / x^-1 y^-1) + (1 / y^-2) ]^(-1/2)



sorry if its confusing (specially the second one...) let me know if you want clarification on any part of it!! thanks for the help!!!


please show steps...i think i have the right answer for 2, but im not sure... i got 1/x + 4/sqrtxy + 1/y




thanks soo much!!!

Answer 1

No...

c^2 = a^2 + b^2
c = 7, a=x b=the distance from the foot of the ladder to the base.

so the distance = sqrt(49-x^2)

Answer 2

Good for you for trying!
First, the ladder problem. This is a right triangle, so the Pythagorean theorem comes into play.
The hypoteneuse is 7 ft
The height of the side opposite the hypoteneuse is x
The equation is 7^2 =x^2 + D^2
D^2=49-x^2
D=rt(49-x^2)

The Pythagorean theorem says H^2=x^2+d^2
You used H^2+x^2=d^2.

1. rt(4x-16) / fourth rt(x-4)^3
rt(4)rt(x-4)/ (x-4)^3/4
2rt(x-4)^1/2 / (x-4)^3/4
2(x-4)^(1/2-3/4)
2(x-4)^-1/4
2/fourth rt(x-4)
Some comments: a^-2=1/a^2
When bases identical you can add exponents for multiplication(a^2 X a^5=a^7) and subtract for division
(a^5/a^2=a^3). In your problem, we had like bases
(x-4)

2. This one I hope I've copied down correctly.
{ 1/x^-2 + 4/(x^-1)(y^-1) + 1/y^-2}^-1/2
{x^2 +4xy +y^2}^-1/2
1/rt(x^2+4xy+y^2)

Expressing your answer in either of the last 2 lines is fully satisfactory.
Some comments: Making 1/x^-2 to be x^2 is fair game for this reason: 1/x^-2 is 1/1 /1/x^2
To divide one fraction by another, we invert the bottom fraction and multiply. [What we are REALLY doing is multiplying 1/1 / 1/x^2 by 1/1. However, we write that 1/1 as x^2/1 / x^2/1. The numerator now becomes 1/1 X x^2/1, or simply x^2. But look what happens to the denominator! 1/x^2 X x^2/1=1!
We end up with x^2 on top, 1 on the bottom]
That's where the rule "invert and multiply" comes from. And that's why what looked like a messy problem got straightened away pretty quickly.

I sincerely hope this rather long explanation is useful to you. Good luck with your math

Answer 3

That's not right.

Let x = height in feet.
Let w = distance from base of latter to wall.

Then:
7 = sqrt(x^2 + w^2)
49 = x^2 + w^2
So:
w^2 = 49 - x^2
w = sqrt(49 - x^2)

For other part:

#1:

sqrt(4x - 16) = sqrt(4*[x-4]) = 2*sqrt(x-4) = 2*(x-4)^(1/2)
fourthroot{(x-4)^3} = {(x-4)^3}^(1/4) = (x-4)^(3*1/4) = (x-4)^(3/4)

So you have:

[ 2*(x-4)^(1/2) ] / (x-4)^(3/4)

Since a^x / a^y = a^(x-y), you have:

2*(x-4)^(1/2 - 3/4) = 2*(x-4)^(2/4-3/4) = 2*(x-4)^(-1/4)

But you want a positive exponent. Since a^x = 1/a^(-x), or conversely, a^(-x) = 1/a^x, you can write it as:

2 / {(x - 4)^(1/4)}

For #2:

1 / x^(-2) = x^2
4 / {x^(-1)y^(-1)} = 4*x^(1)y^(1) = 4xy
1 / y^(-2) = y^2

So it is:

[ x^2 + 4xy + y^2 ]^(-1/2) = 1 / {[ x^2 + 4xy + y^2 ]^(1/2)} = 1/sqrt(x^2 + 4xy + y^2)

Answer 4

well let's see.. 1st one... Think of it as a triangle.
|\
| \
| .\
|__\

Your wall, which is of length "x" is the vertical (dotted) line on the left. Your ladder, which is always 7, is the diagonal one, and the horizontal one is your distance from the base of the wall to the foot of the ladder, your f(x). (ignore the dot, i needed it to make the space)

well the hypotenuse is always 7.
in pythagoras' theory, the squares of the two legs sum up to the square of the hypotenuse. It's usually written as:
a^2 + b^2 = c^2
where "a" and "b" are your two legs and "c" is your hypotenuse.
let's say a is the x feet above the ground, c is your 7-foot ladder. b is your distance from the base or whatever (your f(x)... i'm just gonna put f)

x^2 + f^2 = 49 so solve for f

f^2= 49-x^2 gotta square root both sides
f(x)= sqrt(49-x^2)

OK NEXT ONE.
not even gonna try it. I don't have paper... so sorry. but good luck.

Answer 5

distance = sqrt(49-x^2)

[(4x-16)^1/2]/(x-4)^3/4
=[2(x-4)^1/2]/ (x-4)^3/4
2/(x-4)^1/4

[ (1 / x^-2) + (4 / x^-1 y^-1) + (1 / y^-2) ]^(-1/2)
=[x^2 +4xy +y^2}^-1/2
= 1/(x^2+4xy+y^2) ^1/2

Use these rules:
x^-m = 1/ x^m
and 1/x^-m = x^m

Answer 6

thats right