9. Simplify k ^ 0 * k ^ - 3 . Write the expression using only positive exponents.

Step-by-step explanation:

Purplemath

To simplify with exponents, don't feel like you have to work only with, or straight from, the rules for exponents. It is often simpler to work directly from the definition and meaning of exponents. For instance:

Simplify a6 × a5

The rules tell me to add the exponents. But I when I started algebra, I had trouble keeping the rules straight, so I just thought about what exponents mean. The " a6 " means "six copies of a multiplied together", and the " a5 " means "five copies of a multiplied together". So if I multiply those two expressions together, I will get eleven copies of a multiplied together. That is:

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Simplifying Expressions

a6 × a5 = (a6)(a5)

= ()()

=

= a11

Thus:

a6 × a5 = a11

Simplify the following expression:

\mathbf{\color{green}{\dfrac{6^8}{6^5}}}

6

5

6

8

The exponent rules tell me to subtract the exponents. But let's suppose that I've forgotten the rules again. The " 68 " means I have eight copies of 6 on top; the " 65 " means I have five copies of 6 underneath.

\dfrac{6 \cdot 6 \cdot 6 \cdot 6 \cdot 6 \cdot 6}{6 \cdot 6 \cdot 6 \cdot 6 \cdot 6}

6⋅6⋅6⋅6⋅6

6⋅6⋅6⋅6⋅6⋅6

How many extra 6's do I have, and where are they? I have three extra 6's, and they're on top. Then:

= \dfrac{6 \cdot 6 \cdot 6}{1} = \mathbf{\color{purple}{6^3}}=

1

6⋅6⋅6

=6

3

Unless the instructions also tell you to "evaluate", you're probably expected to leave numerical exponent problems like this in exponent form. If you're not sure, though, feel free to add "= 216", just to be on the safe side.

Simplify the following expression:

\mathbf{\color{green}{\dfrac{\mathit{t}^{10}}{\mathit{t}^8}}}

t

8

t

10

How many extra copies of t do I have, and where are they? I have two extra copies, on top:

\dfrac{t^{10}}{t^8} = \dfrac{t \cdot t \cdot t \cdot t \cdot t \cdot t \cdot t \cdot t \cdot t \cdot t}{t \cdot t \cdot t \cdot t \cdot t \cdot t \cdot t \cdot t}

t

8

t

10

=

t⋅t⋅t⋅t⋅t⋅t⋅t⋅t

t⋅t⋅t⋅t⋅t⋅t⋅t⋅t⋅t⋅t

= \dfrac{t \cdot t}{1} = \mathbf{\color{purple}{\mathit{t}^2}}=

1

t⋅t

=t

2

Once you become comfortable with the "how many extras do I have, and where are they?" reasoning, you'll find yourself not needing to write things out and cancel off the duplicate factors. The answers will start feeling fairly obvious to you.

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Simplify the following expression:

\mathbf{\color{green}{\dfrac{5^3}{5^9}}}

5

9

5

3

This question is a bit different, because the larger exponent is on the term in the denominator. But the basic reasoning is the same.

How many extra copies of 5 do I have, and where are they? I have six extra copies, and they're underneath:

\dfrac{5^3}{5^9} = \dfrac{5 \cdot 5 \cdot 5}{5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5}

5

9

5

3

=

5⋅5⋅5⋅5⋅5⋅5⋅5⋅5⋅5

5⋅5⋅5

= \dfrac{1}{5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5} = \mathbf{\color{purple}{\dfrac{1}{5^6}}}=

5⋅5⋅5⋅5⋅5⋅5

1

=

5

6

1

Note: If you apply the subtraction rule, you'll end up with 53–9 = 5–6, which is mathematically correct, but is almost certainly not the answer they're looking for.

Whether or not you've been taught about negative exponents, when they say "simplify", they mean "simplify the expression so it doesn't have any negative or zero powers". Some students will try to get around this minus-sign problem by arbitrarily switching the sign to magically get " 56 " on top (rather than below a "1"), but this is incorrect.

Let's move on to expressions that are a bit more complex.

Simplify the following expression:

\mathbf{\color{green}{\dfrac{5\mathit{x}^5}{3\mathit{x}^3}}}

3x

3

5x

5

I mustn't forget that the "5" and the "3" are just numbers. Since 3 doesn't go evenly into 5, I can't cancel the numbers.

And I mustn't try to subtract the numbers, because the 5 and the 3 in the fraction "\frac{5}{3}

3

5

" are not at all the same as the 5 and the 3 in rational expression "\frac{x^5}{x^3}

x

3

x

5

". The numerical portion \frac{5}{3}

3

5

 stays as it is.

For the variables, I have two extra copies of x on top, so the answer is:

\dfrac{5x^5}{3x^3} = \dfrac{5 \cdot x \cdot x \cdot x \cdot x \cdot x}{3 \cdot x \cdot x \cdot x}

3x

3

5x

5

=

3⋅x⋅x⋅x

5⋅x⋅x⋅x⋅x⋅x

= \dfrac{5 \cdot x \cdot x}{3}=

3

5⋅x⋅x

= \mathbf{\color{purple}{\dfrac{5\mathit{x}^2}{3} = \dfrac{5}{3}\mathit{x}^2}}=

3

5x

2

=

3

5

x

2

Either of the purple highlighted answers should be acceptable: the only difference is in the formatting; they mean the same thing.

Simplify (–46x2y3z)0

This is simple enough: anything to the zero power is just 1.

(–46x2y3z)0 = 1

Simplify –(46x2y3z)0

The parenthetical portion still simplifies to 1, but this time the "minus" is out in front of the parentheses; that is, it's out from under the power, so the exponent doesn't touch it. So the answer in this case is:

–(46x2y3z)0 = –1

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl