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Gnom [1K]
3 years ago
14

Simplify each expression using the laws of exponents. No negative exponents should be in your final answer. (On the image attach

ed)
WORK NEEDED!

Mathematics
2 answers:
max2010maxim [7]3 years ago
8 0

Answer:

-3x^-3 y^-7 z^8

Step-by-step explanation:

Cloud [144]3 years ago
5 0

Answer:

<em>Answer</em><em> </em><em>is</em><em> </em><em>given below with explanations</em><em>. </em>

Step-by-step explanation:

by \: simplifying \: the \: given \: expression \: using \:  \\ law \: of \: exponents \: we \: get \\  \frac{15 {x}^{15}  {y}^{ - 3}  {z}^{4} }{18 {x}^{18}  { {y}^{4}  {z}^{ - 4} }^{} }  \\   = \frac{5  \: {x}^{ - 3}  {y}^{ - 7} {z}^{8}  }{6}  \\ we \: may \: also \: write \: as \\  =  \frac{5 {z}^{8} }{6 {x}^{3}  {y}^{7} }

formulae \: used  \ \: : \\  \frac{ {x}^{m} }{ { {x}^{n} }^{} }  =  {x}^{m - n}  \:  \:  \: and \\  {x}^{ - m}  =  \frac{1}{ {x}^{m} }

<em>HAVE A NICE DAY</em><em>!</em>

<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>. </em>

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worty [1.4K]

Let's carry this math sentence over to its natural, "shapey" element. We're going to look at each term not as an ordinary number, but as <em>the area of some shape</em>.

x² (read as "x <em>squared"</em>) can be seen as the area of a square with side lengths of x. 2x can similarly be seen as the area of a <em>rectangle </em>with a length of x and a width of 2. (Picture 1)

What's our question actually asking, though? Something about <em>perfect squares</em>. More specifically, we're looking for something to add on that'll <em>make this thing a perfect square</em>. We're trying to find a missing piece we can slot in to make a square, in other words. Problem is, our shapes don't look much like a square if we put them together right now. We need to do a little cutting and gluing first.

First, we're gonna cut the 2x rectangle lengthwise, getting two rectangles with an area of x, a length of 1, and a width of x. Next, we're going to attach them to the x² square, creating this shape that looks, strangely, like a square with a little bit missing from it (picture 2). What we're trying to do is <em>complete this square, </em>to find the area of that little missing chunk.

As it turns out, we have all the information we need for this. Notice that, using the lengths of the x rectangles, we can find that the square's dimensions are 1 x 1, which means that its area is 1 x 1  = 1.

If we tack this new area on to our original expression, we've "completed the square!" We now have a perfect square with side lengths of (x + 1) and an area of (x + 1)² (picture 3).

So, our final expression is x² + 2x + 1, and the missing constant - the area of the "missing square" we had to find to complete our larger one - is 1.

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