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Sever21 [200]
3 years ago
5

If you're ladder is 10 feet long and you place the bottom 6 feet from the wall how far can you reach

Mathematics
1 answer:
-Dominant- [34]3 years ago
6 0
This is like a triangle.
One side, the hypotenuse, is the length of the ladder, 10 feet in this case.
Another side, one of legs, is the distance from the bottom of the ladder to the side of the wall, 6 feet.
The last side is what we need to find, how high up the ladder reaches.

Using the p<span>ythagorean theorem, we can find this third side.
This is written as a^2 + b^2 = c^2.
A and B are the legs, while C is the hypotenuse.

Plugging in known values, we get:
6^2 + b^2 = 10^2

Solve as much as possible:
6^2 = 36
10^2 = 100
36 + b^2 = 100

Now you must isolate b.
Subtract 36 from both sides.
100 - 36 = 64
b^2 = 64

The last step in finding b is doing the inverse of squaring, which is square rooting.
√64 = 8
So b equals 8.

This means that <span>the ladder can reach 8 feet up the wall.</span></span>
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Step-by-step explanation:

\bf \underline{Solution-} \\

We have to simplify the given expression.

\rm =  \dfrac{ {x}^{a + b} \cdot {x}^{b + c} \cdot {x}^{c + a}  }{ {x}^{a} \cdot {x}^{b} \cdot {x}^{c} }

We know that:

\rm \longmapsto {x}^{a} \times  {x}^{b}  =  {x}^{a + b}

\rm \longmapsto \dfrac{ {x}^{a} }{ {x}^{b} } =  {x}^{a -  b}

Therefore, we get:

\rm =  \dfrac{ {x}^{(a + b) + (b + c) + (c + a)}}{ {x}^{a + b + c} }

\rm =  \dfrac{ {x}^{2(a + b+ c)}}{ {x}^{a + b + c} }

\rm = {x}^{2(a + b+ c) - (a + b + c)}

\rm = {x}^{a + b + c}

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