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goldfiish [28.3K]
3 years ago
14

What is 8x times 9x divided by 2xy???

Mathematics
2 answers:
myrzilka [38]3 years ago
7 0

Answer:

\frac{36x}{y}

Step-by-step explanation:

\frac{8x \times 9x}{2xy}  \\ 8 \div 2 \: and \: x \: will \: cancel \: x \: then \: we \: have \\  \frac{4 \times 9x}{y}  \\  \frac{36x}{y}  |final \: answer \: |

alex41 [277]3 years ago
6 0

Answer:

<h2><u>36x /y </u></h2>

Step-by-step explanation:

Hope this helps

<h2><em>:)</em></h2>
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Subtract and it will work
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Kimmy drew the number line below and wrote the comparison shown. Is her comparison correct?
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Step-by-step explanation:

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3 years ago
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Put these in order of smallest first 0.5,55%,3/5
taurus [48]
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55% is converted to 0.55

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5 0
3 years ago
Area of the Region Between Curves-Please help.
Neporo4naja [7]

Answer: 32/3

Step-by-step explanation:

To find the area of the region between the curves, you first want to find the interval of both curves. You can do that by setting both curves equal to each other.

-\frac{1}{2}x^2+2=\frac{1}{2}  x^2-2                   [subtract both sides by 2]

-\frac{1}{2}x^2= \frac{1}{2} x^2-4                         [subtract both sides by \frac{1}{2} x^2]

-x^2=-4\\x=2,-2        

Now that we know the interval, we set this into an integral and subtract the top curve by the bottom curve. *Note: I can't type in -2 into the interval, therefore only a "-" will be displayed.

\int\limits^2_- {|-\frac{1}{2}x^2+2-(\frac{1}{2}x^2-2) | } \, dx                     [distribute -1]

\int\limits^2_-{|-\frac{1}{2}x^2+2-\frac{1}{2}x^2+2  |} \, dx                       [combine like terms]

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\frac{32}{3}

The area between the curves is 32/3. This was also the reason why we had the absolute values because <u>area can never be negative</u>.

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Answer:

D

Step-by-step explanation:

3 0
3 years ago
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