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Advocard [28]
3 years ago
12

7B2%7D%20%7D%7B%20%5Cfrac%7B1%7D%7By%7D%20%7D%20" id="TexFormula1" title=" \frac{ {3x}^{2} }{xy} \times \frac{ {4xy}^{2} }{ \frac{1}{y} } " alt=" \frac{ {3x}^{2} }{xy} \times \frac{ {4xy}^{2} }{ \frac{1}{y} } " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
hjlf3 years ago
6 0
\bf ~~~~~~~~~~~~\textit{negative exponents}
\\\\
a^{-n} \implies \cfrac{1}{a^n}
\qquad \qquad
\cfrac{1}{a^n}\implies a^{-n}
\qquad \qquad 
a^n\implies \cfrac{1}{a^{-n}}
\\\\
-------------------------------\\\\
\cfrac{3x^2}{xy}\cdot \cfrac{4xy^2}{\frac{1}{y}}\implies \cfrac{3x^2}{xy}\cdot \cfrac{4xy^2}{y^{-1}}\implies \cfrac{3x^2x^1y^2}{xy^1y^{-1}}\implies \cfrac{3x^{2+1}y^2}{xy^{1-1}}
\\\\\\
\cfrac{3x^3y^2}{xy^0}\implies \cfrac{3x^3y^2}{x^1}\implies 3x^3y^2x^{-1}\implies 3x^{3-1}y^2\implies 3x^2y^2
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Can anyone solve and or help me with this?
Ad libitum [116K]

18 = 2x + 4

2x = 14

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answer

first line:  1st box: 18 and 2nd box: 4

second line:  $7 an hour

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4 years ago
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photoshop1234 [79]

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3 years ago
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3 years ago
Read 2 more answers
A charitable organization packed 72 bags of rice, 48 blankets, and 24 cases of water equally in the boxes. Find the greatest pos
AysviL [449]

Answer:

The greatest possible number of boxes that is needed to pack the items with no leftovers is 24.

Step-by-step explanation:

Given:

Number of rice bags = 72

Number of blankets =48

Number of water cases = 24

To Find :

The greatest possible number of boxes that the items can be packed into so that there are no leftover = ?

Solution:

we can find the greatest possible number of boxes that is required by finding the HCF(Highest common factor) of(24,48,72)

Finding the HCF (Highest common factor):

The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24

The factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

24 is the greatest common factor

Hence 24 will the number boxes required

6 0
3 years ago
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